Apply linear & quadratic equations

1. Revision of solving equations

Exercises

2. Revision of simultaneous equations

Points of intersection

Draw the graphs of the following equations on the same Cartesian plane. Use the graphs to find the points of intersection of the two lines.

\begin{aligned} y & = x^{2} + 4 x + 3 \\ y & = 2 x + 6 \end{aligned}

$\begin{aligned} y & = x^{2} + 4 x + 3 \\ y & = 2 x + 6 \end{aligned}$

INSTRUCTION:

You should get two coordinate pairs in the form $\left(x_1; y_1\right)$ and $\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The points of intersection are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to plot the graphs of the two equations on the same Cartesian plane.

STEP: Plot the graph of

y = x^{2} + 4 x + 3

$y = x^{2} + 4 x + 3$ on a Cartesian plane
[−2 points ⇒ 4 / 6 points left]

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: $y = x^{2} + 4 x + 3$ .

To plot this graph, we need to find the coordinates of the points where the graph will cross the $x$ - and $y$ -axes. These points are the $x$ - and $y$ -intercepts, respectively. We will then use these points to plot the graph.

The $y$ -intercept is the quantity in the equation which is not multiplied by $x$ . For this equation, the $y$ -intercept is $\text{3}$ . The coordinates of this point are $\left(0; 3\right)$ .

The $x$ -intercepts are the roots of the equation $x^{2} + 4 x + 3 = 0$ . We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{4}$ . This is the coefficient of the middle term. The product of the two numbers must be $\text{3}$ . This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{3}$ . The two numbers are $\text{1}$ and $\text{3}$ . We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = x^{2} + 4 x + 3 \\ 0 & = x^{2} + x + 3 x + 3 \end{aligned}

$\begin{aligned} 0 &=x^{2} + 4 x + 3 \\ 0 &= x^{2}+x+3 x+3 \end{aligned}$

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

\begin{aligned} 0 & = x (x + 1) + 3 (x + 1) \\ 0 & = (x + 3) (x + 1) \end{aligned}

$\begin{aligned} 0 &= x(x+1)+3(x+1)\\ 0 &= (x+3)(x+1) \end{aligned}$

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are $x = -1$ and $x = -3$ . The coordinates of these roots are $\left(-1; 0 \right)$ and $\left(-3; 0 \right)$ .

Using these $x$ - and $y$ -intercepts we will draw the graph:

STEP: Plot the graph of

y = 2 x + 6

$y = 2 x + 6$ on the same Cartesian plane
[−2 points ⇒ 2 / 6 points left]

As we did for the quadratic equation, we will first find the $y$ -intercept. After that we will calculate root for the linear equation. We remember that the $y$ -intercept is the quantity in the equation which is not multiplied by $x$ .

The $y$ -intercept is $\text{6}$ . Its coordinates are: $\left(0; 6\right)$ .

We will work out the root of the equation $0 = 2 x + 6$ .

\begin{aligned} 0 & = 2 x + 6 & = 0 \\ 2 x & = - 6 \\ x & = - 3 \end{aligned}

$\begin{aligned} 0 & = 2 x + 6 &= 0\\ 2 x &= -6\\x &= -3 \end{aligned}$

The root is: $x = -3$ . It's coordinates are: $\left( -3; 0 \right)$ .

We will now plot the equation $y = 2 x + 6$ on the same Cartesian plane we have used for the graph of the quadratic equation.

STEP: Read off the coordinates of the points of intersection
[−2 points ⇒ 0 / 6 points left]

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: $(1; 8)$ and $(-3; 0)$ .

We have used a graphical method to determine the solutions of the simultaneous equations: $y = x^{2} + 4 x + 3$ and $y = 2 x + 6$ . The solutions are the points where their graphs intersect.

NOTE: It is possible to solve the simultaneous equations using either the substitution or elimination methods.

The points of intersection are $(1; 8)$ and $(-3; 0)$ .

Submit your answer as: and

Simultaneous equations: solving by substitution

Solve simultaneously for $x$ and $y$ . You should get two coordinate pairs, $(x;y)$ , for your answers.

\begin{aligned} y + 4 & = - x - 18 \\ x^{2} + y + 16 & = 0 \end{aligned}

$\begin{aligned} y + 4 & = - x - 18 \\ x^{2} + y + 16 & = 0 \end{aligned}$

INSTRUCTION: When you type your answers they should look something like this: ( 2; -4 ) and ( -1; 3). It does not matter which coordinate pair you type first.

Answer: The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.

STEP: Rearrange the linear equation to isolate a variable
[−1 point ⇒ 5 / 6 points left]

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate $y$ .

\begin{aligned} y + 4 & = - x - 18 \\ y & = - x - 22 \end{aligned}

$\begin{aligned} y + 4 & = - x - 18 \\ y & = - x - 22 \end{aligned}$

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 6 points left]

Now we need to substitute the entire right side of the equation, $(- x - 22)$ , into the quadratic equation in place of $y$ . Be sure to use brackets! Then simplify the equation to get standard form.

\begin{aligned} x^{2} + y + 16 & = 0 \\ x^{2} + (- x - 22) + 16 & = 0 \\ x^{2} - x - 6 & = 0 \end{aligned}

$\begin{aligned} x^{2} + y + 16 & = 0 \\ x^{2} + \left(- x - 22\right) + 16 & = 0 \\ x^{2} - x - 6 & = 0 \end{aligned}$

STEP: Solve for the values of

x

$x$
[−1 point ⇒ 2 / 6 points left]

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two $x$ -values.

\begin{aligned} x^{2} - x - 6 & = 0 \\ (x - 3) (x + 2) & = 0 \\ x = 3 & and x = - 2 \end{aligned}

$\begin{aligned} x^{2} - x - 6 & = 0 \\ \left(x - 3\right) \left(x + 2\right) & = 0 \\ x = 3 \quad & \text{ and } \quad x = -2 \end{aligned}$

STEP: Find the value of

y

$y$ for each of the

x

$x$ -values
[−2 points ⇒ 0 / 6 points left]

Cool! However, there is still work to do. We must now find the values of $y$ for these equations. We must substitute in the values $x = 3$ and $x = -2$ . We can use either equation to do this, so we will choose the easier option: the linear equation.

\begin{aligned} If x = 3 : \\ y & = - (3) - 22 \\ y & = - 25 \\ If x = - 2 : \\ y & = - (- 2) - 22 \\ y & = - 20 \end{aligned}

$\begin{aligned} \text{If } \; x = 3: \; & \\ y & = - (3) - 22 \\ y & = -25 \\ \text{If } \; x = -2: \; & \\ y & = - (-2) - 22 \\ y & = -20 \end{aligned}$

Finally we have the answers: if $x = 3$ then $y = -25$ and if $x = -2$ then $y = -20$ .

NOTE: The graph below is not required for the solution. It is here to show the connection between the answers and the equations.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: $\left(3 ; -25\right)$ and $\left(-2 ; -20\right)$ .

Submit your answer as: and

Simultaneous equations

Solve for $x$ and $y$ from the given equations:

\begin{aligned} y & = x^{2} + 2 x \\ y & = x + 2 \end{aligned}

$\begin{aligned} y & = x^{2} + 2 x \\ y & = x + 2 \end{aligned}$

INSTRUCTION: You should get two coordinate pairs in the form

(x_{1}; y_{1})

$\left(x_1; y_1\right)$ and

(x_{2}; y_{2})

$\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided. It does not matter which coordinate pair you type first.

Answer:

The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting $x + 2$ in place of $y$ in the quadratic equation and then simplify.

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

\begin{aligned} quadratic equation ⟶ & y = x^{2} + 2 x \\ linear equation ⟶ & y = x + 2 \end{aligned}

$\begin{aligned} {\text{quadratic equation}} \quad \longrightarrow \quad & y = x^{2} + 2 x \\ {\text{linear equation}} \quad \longrightarrow \quad & y = x + 2 \end{aligned}$

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

\begin{aligned} x + 2 & = x^{2} + 2 x \\ 0 & = x^{2} + 2 x - x - 2 \\ 0 & = x^{2} + 2 x - x - 2 \\ 0 & = x^{2} + x - 2 \end{aligned}

$\begin{aligned} x + 2 &= x^{2} + 2 x \\ 0 &= x^{2} + 2 x-x-2 \\ 0 &= x^{2}+2 x-x-2\\ 0 &= x^{2}+x-2 \end{aligned}$

STEP: Solve the equation

x^{2} + x - 2 = 0

$x^{2}+x-2 = 0$
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{1}$ . This is the coefficient of the middle term. The product of the two numbers must be $-\text{2}$ . This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $-\text{2}$ . The two numbers are $\text{2}$ and $-\text{1}$ . We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = x^{2} + x - 2 \\ 0 & = x^{2} + 2 x - x - 2 \end{aligned}

$\begin{aligned} 0 &= x^{2}+x-2 \\ 0 &= x^{2}+2 x-x-2 \end{aligned}$

We now group the first two terms together and group the last terms together and factorise.

\begin{aligned} 0 & = x^{2} + 2 x - x - 2 \\ 0 & = x (x + 2) - (x + 2) \end{aligned}

$\begin{aligned} 0 &= \color{red}{x^{2}+2 x}\color{blue}{-x-2}\\ 0 &= \color{red}{x(x+2)}\color{blue}{-(x+2)} \end{aligned}$

The last thing now is to factor the highest common factor.

\begin{aligned} 0 & = (x - 1) (x + 2) \end{aligned}

$\begin{aligned} 0 &= (x-1)(x+2) \end{aligned}$

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$ -values of our solutions. The $x$ -values are $\text{1}$ and $-\text{2}$ .

STEP: Calculate the corresponding values of

y

$y$
[−2 points ⇒ 0 / 6 points left]

We will substitute each $x$ -value into the linear equation. This will give us the corresponding values of $y$ .

For $x = 1$ :

\begin{aligned} y & = x + 2 \\ = 3 \end{aligned}

$\begin{aligned} y &= x + 2 \\ &= 3 \end{aligned}$

For $x = -2$ :

\begin{aligned} y & = x + 2 \\ = 0 \end{aligned}

$\begin{aligned} y &= x + 2 \\ &= 0 \end{aligned}$

The solutions are $x = 1$ together with $y = 3$ , and $x = -2$ with $y = 0$ . In coordinates pairs, the solutions are $\left(1; 3\right)$ and $\left(-2; 0\right)$ .

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(1; 3\right)$ and $\left(-2; 0\right)$ . These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(1; 3\right)$ and $\left(-2; 0\right)$ .

Submit your answer as: and

Simultaneous equations

Determine the solution of the following equations:

\begin{aligned} y & = 2 x^{2} + 8 x + 7 \\ y & = x + 4 \end{aligned}

$\begin{aligned} y & = 2 x^{2} + 8 x + 7 \\ y & = x + 4 \end{aligned}$

INSTRUCTION:

You should get two coordinate pairs in the form $\left(x_1; y_1\right)$ and $\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting $x + 4$ in place of $y$ in the quadratic equation and then simplify.

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

\begin{aligned} quadratic equation ⟶ & y = 2 x^{2} + 8 x + 7 \\ linear equation ⟶ & y = x + 4 \end{aligned}

$\begin{aligned} {\text{quadratic equation}} \quad \longrightarrow \quad & y = 2 x^{2} + 8 x + 7 \\ {\text{linear equation}} \quad \longrightarrow \quad & y = x + 4 \end{aligned}$

This means we want two pairs of $x$ and $y$ -values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of

y

$y$ from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.

\begin{aligned} x + 4 & = 2 x^{2} + 8 x + 7 \\ 0 & = 2 x^{2} + 8 x + 7 - x - 4 \\ 0 & = 2 x^{2} + 8 x - x + 7 - 4 \\ 0 & = 2 x^{2} + 7 x + 3 \end{aligned}

$\begin{aligned} x + 4 &= 2 x^{2} + 8 x + 7 \\ 0 &= 2 x^{2} + 8 x + 7-x-4 \\ 0 &= 2 x^{2}+8 x-x+7-4\\ 0 &= 2 x^{2}+7 x+3 \end{aligned}$

STEP: Solve the equation

0 = 2 x^{2} + 7 x + 3

$0 = 2 x^{2}+7 x+3$
[−2 points ⇒ 2 / 6 points left]

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{7}$ . This is the coefficient of the middle term. The product of the two numbers must be $\text{6}$ . This is the product of the coefficient of the first term $\text{2}$ and the value of the last term $\text{3}$ . The two numbers are $\text{1}$ and $\text{6}$ . We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = 2 x^{2} + 7 x + 3 \\ 0 & = 2 x^{2} + x + 6 x + 3 \end{aligned}

$\begin{aligned} 0 &= 2 x^{2}+7 x+3 \\ 0 &= 2 x^{2}+x+6 x+3 \end{aligned}$

We now group the first two terms together and group the last terms together and factorise.

\begin{aligned} 0 & = 2 x^{2} + x + 6 x + 3 \\ 0 & = x (2 x + 1) + 3 (2 x + 1) \end{aligned}

$\begin{aligned} 0 &= \color{red}{2 x^{2}+x}\color{blue}{+6 x+3}\\ 0 &= \color{red}{x(2 x+1)}\color{blue}{+3(2 x+1)} \end{aligned}$

The last thing now is to factor the highest common factor.

\begin{aligned} 0 & = (x + 3) (2 x + 1) \end{aligned}

$\begin{aligned} 0 &= (x+3)(2 x+1) \end{aligned}$

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$ -values of our solutions. The $x$ -values are $-\dfrac{1}{2}$ and $-\text{3}$ .

STEP: Calculate the corresponding values of

y

$y$
[−2 points ⇒ 0 / 6 points left]

We will substitute each $x$ -value into the linear equation. This will give us the corresponding values of $y$ .

NOTE: We can do that by substituting

x

$x$ into any of the original equations. It is easier to use the linear one.

For $x = -\dfrac{1}{2}$ :

\begin{aligned} y & = x + 4 \\ = \frac{7}{2} \end{aligned}

$\begin{aligned} y &= x + 4 \\ &= \dfrac{7}{2} \end{aligned}$

For $x = -3$ :

\begin{aligned} y & = x + 4 \\ = 1 \end{aligned}

$\begin{aligned} y &= x + 4 \\ &= 1 \end{aligned}$

The solutions are $x = -\dfrac{1}{2}$ together with $y = \dfrac{7}{2}$ , and $x = -3$ with $y = 1$ . In coordinate pairs, the solutions are $\left(-\dfrac{1}{2}; \dfrac{7}{2}\right)$ and $\left(-3; 1\right)$ .

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(-\dfrac{1}{2}; \dfrac{7}{2}\right)$ and $\left(-3; 1\right)$ . These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(-\dfrac{1}{2}; \dfrac{7}{2}\right)$ and $\left(-3; 1\right)$ .

Submit your answer as: and

Simultaneous equations

Solve for $x$ and $y$ :

\begin{aligned} y & = x^{2} + 2 x - 1 \\ x & = y - 1 \end{aligned}

$\begin{aligned} y & = x^{2} + 2 x - 1 \\ x & = y - 1 \end{aligned}$

INSTRUCTION:

You should get two coordinate pairs in the form $\left(x_1; y_1\right)$ and $\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting $y - 1$ in place of $x$ in the quadratic equation and then simplify.

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

\begin{aligned} quadratic equation ⟶ & y = x^{2} + 2 x - 1 \\ linear equation ⟶ & x = y - 1 \end{aligned}

$\begin{aligned} {\text{quadratic equation}} \quad \longrightarrow \quad & y = x^{2} + 2 x - 1 \\ {\text{linear equation}} \quad \longrightarrow \quad & x = y - 1 \end{aligned}$

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $x$ from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of

y

$y$ from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.

\begin{aligned} y & = {(y - 1)}^{2} + 2 (y - 1) - 1 \\ 0 & = {(y - 1)}^{2} + 2 (y - 1) - 1 - y \\ 0 & = y^{2} - 2 y + 1 + (2 y - 2) - 1 - y \\ 0 & = y^{2} - 2 y + 1 + 2 y - 2 - 1 - y \\ 0 & = y^{2} - 2 y + 2 y - y + 1 - 2 - 1 \\ 0 & = y^{2} - y + 1 - 2 - 1 \\ 0 & = y^{2} - y - 1 - 1 \\ 0 & = y^{2} - y - 2 \end{aligned}

$\begin{aligned} y &= \left(y - 1\right)^2+2\left(y - 1\right)-1 \\ 0 &= {{{{{\left({{y}-{1}}\right)}^{2}}+{{2}\,{}{\left({{y}-{1}}\right)}}}-{1}}-{y}}\\0 &= {{{{{{{y}^{2}}-{{2}\,{}{y}}}+{1}}+{\left({{{2}\,{}{y}}-{2}}\right)}}-{1}}-{y}}\\0 &= {{{{{{{{y}^{2}}-{{2}\,{}{y}}}+{1}}+{{2}\,{}{y}}}-{2}}-{1}}-{y}}\\0 &= {{{{{{{{y}^{2}}-{{2}\,{}{y}}}+{{2}\,{}{y}}}-{y}}+{1}}-{2}}-{1}}\\0 &= {{{{{{y}^{2}}-{y}}+{1}}-{2}}-{1}}\\0 &= {{{{{y}^{2}}-{y}}-{1}}-{1}}\\0 &= {{{{y}^{2}}-{y}}-{2}} \end{aligned}$

STEP: Solve the equation

0 = y^{2} - y - 2

$0 = {{{{y}^{2}}-{y}}-{2}}$
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{1}$ . This is the coefficient of the middle term. The product of the two numbers must be $-\text{2}$ . This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $-\text{2}$ . The two numbers are $\text{1}$ and $-\text{2}$ . We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = y^{2} - y - 2 \\ 0 & = y^{2} + y - 2 y - 2 \end{aligned}

$\begin{aligned} 0 &= {{{{y}^{2}}-{y}}-{2}} \\ 0 &= y^{2}+y-2 y-2 \end{aligned}$

We now group the first two terms together and group the last terms together and factorise.

\begin{aligned} 0 & = y^{2} + y - 2 y - 2 \\ 0 & = y (y + 1) - 2 (y + 1) \end{aligned}

$\begin{aligned} 0 &= \color{red}{y^{2}+y}\color{blue}{-2 y-2}\\ 0 &= \color{red}{y(y+1)}\color{blue}{-2(y+1)} \end{aligned}$

The last thing now is to factor the highest common factor.

\begin{aligned} 0 & = (y - 2) (y + 1) \end{aligned}

$\begin{aligned} 0 &= (y-2)(y+1) \end{aligned}$

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $y$ -values of our solutions. The $y$ -values are 2 and -1.

STEP: Calculate the corresponding values of

x

$x$
[−2 points ⇒ 0 / 6 points left]

We will substitute each $y$ -value into the linear equation. This will give us the corresponding values of $x$ .

NOTE: We can do that by substituting

y

$y$ into any of the original equations. It is easier to use the linear one.

For $y = 2$ :

\begin{aligned} x & = y - 1 \\ = 1 \end{aligned}

$\begin{aligned} x &= y - 1 \\ &= 1 \end{aligned}$

For $y = -1$ :

\begin{aligned} x & = y - 1 \\ = - 2 \end{aligned}

$\begin{aligned} x &= y - 1 \\ &= -2 \end{aligned}$

The solutions are $x = 1$ together with $y = 2$ , and $x = -2$ with $y = -1$ . In coordinates pairs, the solutions are $\left(1; 2\right)$ and $\left(-2; -1\right)$ .

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(1; 2\right)$ and $\left(-2; -1\right)$ . These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(1; 2\right)$ and $\left(-2; -1\right)$ .

Submit your answer as: and

Simultaneous equations: intersection points

The graph below shows the equations:

\begin{matrix} y = \frac{x}{2} - 3 and y = \frac{x^{2}}{2} + x - 4 \end{matrix}

$\begin{array}{c}y = \frac{x}{2} - 3 \quad \text{and} \quad y = \frac{x^{2}}{2} + x - 4\end{array}$

Determine the points of intersection for these two curves.

INSTRUCTION:

Write your answers as coordinate pairs like this: ( 3; -4) and ( 1; 8).
It does not matter which coordinate pair you type first.
Do not round your answers.

Answer: The points of intersection are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Solve the equations simultaneously to find their intersection points.

STEP: Combine the equations by substitution
[−1 point ⇒ 5 / 6 points left]

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation $y$ is equal to the quantity $\frac{x}{2} - 3$ ; and at the same time, the quadratic equation also contains the value $y$ . Substitute like this:

\begin{aligned} y & is \frac{x}{2} - 3 \\ and: \\ y & is also \frac{x^{2}}{2} + x - 4 \\ Therefore: \\ \frac{x}{2} - 3 & = \frac{x^{2}}{2} + x - 4 \end{aligned}

$\begin{aligned} y & \; \text{ is } \; \frac{x}{2} - 3 \\ \text{and:} \quad & \\ y & \; \text {is also } \; \frac{x^{2}}{2} + x - 4 \\ \text{Therefore: } \quad & \\ \frac{x}{2} - 3 & = \frac{x^{2}}{2} + x - 4 \end{aligned}$

STEP: Arrange the equation in standard form
[−2 points ⇒ 3 / 6 points left]

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is $\text{2}$ . So multiply both sides of the equation by $\text{2}$ in order to cancel the denominators. Then arrange the equation in standard form.

\begin{aligned} (2) (\frac{x}{2} - 3) & = (\frac{x^{2}}{2} + x - 4) (2) \\ x - 6 & = x^{2} + 2 x - 8 \\ 0 & = x^{2} + x - 2 \end{aligned}

$\begin{aligned} (\text{2})\left(\frac{x}{2} - 3\right) & = \left(\frac{x^{2}}{2} + x - 4\right)(\text{2}) \\ x - 6 & = x^{2} + 2 x - 8 \\ 0 & = x^{2} + x - 2 \end{aligned}$

STEP: Solve the equation
[−1 point ⇒ 2 / 6 points left]

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two $x$ -values.

\begin{aligned} 0 & = x^{2} + x - 2 \\ 0 & = (x - 1) (x + 2) \\ x = - 2 & and x = 1 \end{aligned}

$\begin{aligned} 0 & = x^{2} + x - 2 \\ 0 & = \left(x - 1\right) \left(x + 2\right) \\ x = -2 & \quad \textrm{ and } \quad x = 1 \end{aligned}$

STEP: Use the values of

x

$x$ to find the values of

y

$y$
[−2 points ⇒ 0 / 6 points left]

Great... but it isn't over yet. We must now find the values of $y$ for these equations. Substitute in the values $x = -2$ and $x = 1$ into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

\begin{aligned} If x = - 2 : \\ y & = \frac{(- 2)}{2} - 3 \\ y & = - 4 \\ If x = 1 : \\ y & = \frac{(1)}{2} - 3 \\ y & = - \frac{5}{2} \end{aligned}

$\begin{aligned} \text{If } \; x = -2: \; & \\ y & = \frac{(-2)}{2} - 3 \\ y & = -4 \\ \text{If } \; x = 1: \; & \\ y & = \frac{(1)}{2} - 3 \\ y & = - \frac{5}{2} \end{aligned}$

Finally! The solutions are $\left(-2 ; -4\right)$ and $\left(1 ; - \frac{5}{2}\right)$ .

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are $\left(-2 ; -4\right)$ and $\left(1 ; - \frac{5}{2}\right)$ .

Submit your answer as: and

Substitution with quadratic equations

Here are two equations, which we can solve simultaneously using substitution:

\begin{aligned} y & = x^{2} + 2 x - 1 \\ x & = y - 1 \end{aligned}

$\begin{aligned} y &= x^{2} + 2 x - 1 \\ x & = y - 1 \end{aligned}$

Which of the following equations shows the correct substitution step?

A	$y=(- y + 1)^2+2(- y + 1)-1$
B	$y=(y - 1)^2+2(y - 1)-1$
C	$y=y^{2} + 2 y - 1$
D	$y=y-1^2+2 y-1-1$

Answer: The correct substitution step is choice: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to substitute $y - 1$ in place of $x$ in the quadratic equation $y = x^{2} + 2 x - 1$ .

STEP: Substitute the linear equation into the quadratic equation
[−1 point ⇒ 0 / 1 points left]

We have two equations that we will solve by substitution. The equations are:

\begin{aligned} quadratic equation ⟶ & y = x^{2} + 2 x - 1 \\ linear equation ⟶ & x = y - 1 \end{aligned}

$\begin{aligned} {\text{quadratic equation}} \quad \longrightarrow \quad & y = x^{2} + 2 x - 1 \\ {\text{linear equation}} \quad \longrightarrow \quad & x = y - 1 \end{aligned}$

We need to substitute $y - 1$ from the linear equation in place of $x$ in the quadratic equation.

\begin{matrix} y = (y - 1)^{2} + 2 (y - 1) - 1 \end{matrix}

$\begin{array}{c}y=(\color{red}{y - 1})^2+2(\color{red}{y - 1})-1\end{array}$

NOTE: We could also substitute the value of

y

$y$ from the quadratic equation into the linear equation. The solution for this question did not consider this case.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for $y$ . The question does not ask us to solve the simultaneous equations.

The correct substitution step is $y=(y - 1)^2+2(y - 1)-1$ which is choice B.

Submit your answer as:

Equations with fractions

Determine the values of $x$ and $y$ which solve these two equations:

\begin{aligned} 3 x - y & = - 10 \\ - \frac{3}{x} & = - \frac{4}{y} - \frac{20}{x y} x, y \neq 0 \end{aligned}

$\begin{aligned} 3 x - y &= -10 \\ - \frac{3}{x} &= - \frac{4}{y} - \frac{20}{x y} \qquad x, y \neq 0 \end{aligned}$

Your answer should be exact (do not round off).

Answer:

The solution is $x =$ and $y =$ .

numeric

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The first thing you should do is rearrange the second equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is $xy$ .

STEP: Rearrange the second equation
[−1 point ⇒ 4 / 5 points left]

We need to solve the equations simultaneously. To do that, we should start by rearranging the second equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by $xy$ , which is the LCD of the fractions. That will cancel all of the denominators.

\begin{aligned} - \frac{3}{x} & = - \frac{4}{y} - \frac{20}{x y} \\ x y (- \frac{3}{x}) & = x y (- \frac{4}{y} - \frac{20}{x y}) \\ - \frac{3}{x} x y & = - \frac{4}{y} x y - 20 \\ - 3 y & = - 4 x - 20 \end{aligned}

$\begin{aligned} - \frac{3}{x} &= - \frac{4}{y} - \frac{20}{x y} \\ x y\left(- \frac{3}{x}\right)&=x y\left(- \frac{4}{y} - \frac{20}{x y}\right) \\ - \frac{3}{x} x y &= - \frac{4}{y} x y - 20 \\ - 3 y &= - 4 x - 20 \end{aligned}$

STEP: Pick a method and solve for whichever variable comes more easily
[−3 points ⇒ 1 / 5 points left]

Now we are ready to solve the following equations:

\begin{aligned} 3 x - y = - 10 \\ - 3 y = - 4 x - 20 \end{aligned}

$\begin{aligned} 3 x - y = -10 \\ - 3 y = - 4 x - 20 \end{aligned}$

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations are not arranged nicely for elimination or substitution. We must change the equations somehow. In this case, we can make a small change that will allow us to use elimination: we can multiply the first equation by $\text{3}$ so that the $y$ -coefficients will be ready to cancel. Then we can eliminate them by subtracting the equations.

First modify things to set up the elimination:

\begin{aligned} 3 (3 x - y) & = 3 (- 10) \\ 9 x - 3 y & = - 30 \end{aligned}

$\begin{aligned} 3 \left(3 x - y \right) & = 3 \left(-10 \right) \\ 9 x - 3 y & = -30 \end{aligned}$

Now do the elimination and complete the solution to find $x$ .

\begin{aligned} 9 x - 3 y & = - 30 \\ _{second equation}^{Subtract the} \to - (- 3 y) & = - (- 4 x - 20) \\ 9 x - 3 y + 3 y & = - 30 + 20 + 4 x \\ 9 x & = 4 x - 10 \\ 5 x & = - 10 \\ x & = - 2 \end{aligned}

$\begin{aligned} 9 x - 3 y & = -30 \\ ^{\text{Subtract the}}_{\text{second equation}} \quad \rightarrow \quad -\left(- 3 y\right)& =-\left(- 4 x - 20\right) \\ \hline 9 x - 3 y + 3 y & = - 30 + 20 + 4 x \\ 9 x & = 4 x - 10 \\ 5 x & = -10 \\ x & = -2 \end{aligned}$

Great - we have the first value, $x = -2$ .

STEP: Solve for the other variable
[−1 point ⇒ 0 / 5 points left]

Now the last step: find the value of $y$ . We can do this using either of the equations in the question: but the first eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the second equation).

\begin{aligned} 3 x - y & = - 10 \\ 3 (- 2) - y & = - 10 \\ - y & = - 4 \\ y & = 4 \end{aligned}

$\begin{aligned} 3 x - y & = -10 \\ 3 \left(-2\right) - y & = -10 \\ - y & = -4 \\ y & = 4 \end{aligned}$

The answer is the pair of numbers $x = -2$ and $y = 4$ . As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

NOTE: Remember, the fractions force

x, y \neq 0

$x, y \neq 0$ , which is why there are open intervals at the

x

$x$ - and

y

$y$ -intercepts of the fraction equation.

The values which solve these two equations are $x = -2$ and $y = 4$ .

Submit your answer as: and

Simultaneous equations: special outcomes

Solve for $x$ and $y$ from the given equations:

\begin{aligned} y & = 3 x^{2} + 3 x + 2 \\ y & = - x - 5 \end{aligned}

$\begin{aligned} y & = 3 x^{2} + 3 x + 2 \\ y & = - x - 5 \end{aligned}$

INSTRUCTION: You should enter your answer in the form of a coordinate pair, (x; y). An example of an acceptable answer is (2; 5). If there is no solution, type no solution.

Answer: The solution is .

one-of

type(string.nocase)

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to start by substituting $- x - 5$ in place of $y$ in the quadratic equation and simplify. You must expect a special kind of solution.

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 2 / 4 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is $y = 3 x^{2} + 3 x + 2$ and the linear equation is $y = - x - 5$ . This means we expect two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

NOTE: We could also substitute the value of

y

$y$ from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.

\begin{aligned} - x - 5 & = 3 x^{2} + 3 x + 2 \\ 0 & = 3 x^{2} + 3 x + 2 + x + 5 \\ 0 & = 3 x^{2} + 3 x + x + 2 + 5 \\ 0 & = 3 x^{2} + 4 x + 7 \end{aligned}

$\begin{aligned} - x - 5 &= 3 x^{2} + 3 x + 2 \\ 0 &= 3 x^{2} + 3 x + 2+x+5\\ 0 &= 3 x^{2}+3 x+x+2+5\\ 0 &= 3 x^{2}+4 x+7 \end{aligned}$

STEP: Solve the equation

0 = 3 x^{2} + 4 x + 7

$0=3 x^{2}+4 x+7$
[−2 points ⇒ 0 / 4 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

We need two numbers that add to give $\text{4}$ . This is the coefficient of the middle term. The product of the two numbers must be $\text{21}$ . This is the product of the coefficient of the first term $\text{3}$ and the value of the last term $\text{7}$ . In this case, there are no such numbers. We will now use the quadratic formula to solve this equation.

The solution for a quadratic equation of the form $ax^2 + bx + c = 0$ :

\begin{matrix} x = \frac{- b \pm \sqrt{(b^{2} - 4 a c)}}{2 a} \end{matrix}

$\begin{array}{c}x = \dfrac{-b \pm \sqrt{\left(b^2 - 4ac\right)}}{2a}\end{array}$

Before we can use the quadratic formula, we need to identify the coefficients $a$ , $b$ and $c$ from the equation $3 x^{2}+4 x+7 = 0$ These are:

\begin{matrix} a = 3, b = 4 and c = 7. \end{matrix}

$\begin{array}{c}a = 3, b = 4 \text{ and } c = 7.\end{array}$

We will substitute these values into the formula and simplify.

\begin{aligned} x & = \frac{- (4) \pm \sqrt{({(4)}^{2} - 4 (3) (7))}}{2 (3)} \\ = \frac{- 4 \pm \sqrt{(16 - (84))}}{6} \\ = \frac{- 4 \pm \sqrt{(- 68)}}{6} \end{aligned}

$\begin{aligned} x &= \dfrac{-\left(4\right) \pm \sqrt{\left(\left(4\right)^2 - 4\left(3\right) \left(7\right)\right)}} {2\left(3\right)} \\ &= \dfrac{-4 \pm \sqrt{\left(16 - \left(84\right)\right)}} {6} \\ &= \dfrac{-4 \pm \sqrt{\left(-68\right)}} {6} \end{aligned}$

The number under the square root sign is negative. This means that our quadratic equation has no real solution. This means that $y = 3 x^{2} + 3 x + 2$ and $y = - x - 5$ cannot be solved simultaneously.

We can confirm our result by plotting the graph of each equation on the same set of axes. We do not expect these graphs to intersect.

Aha! The two graphs do not have an intersection point. This means that indeed the two equations have no solution.

The final answer is no solution.

Submit your answer as:

Simultaneous equations

Adapted from DBE Nov 2016 Grade 12, P1, Q1.3
Maths formulas

Solve for $x$ and $y$ :

\begin{matrix} x = - 2 y \\ and \\ x^{2} + 3 x y = - 18 \end{matrix}

$\begin{array}{c} x = - 2 y \\ \text{and} \\ x^{2} + 3 x y= -18 \end{array}$

INSTRUCTIONS:

You should get two coordinate pairs in the form $(x_{1}; y_{1})$ and $(x_{2}; y_{2})$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The solutions are
and .

coordinate

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.

STEP: Substitute

x = - 2 y

$x = - 2 y$ into

x^{2} + 3 x y = - 18

$x^{2} + 3 x y= -18$ and solve for

y

$y$
[−3 points ⇒ 1 / 4 points left]

We need to solve the following pair of equations for $x$ and $y$ :

\begin{matrix} x = - 2 y \\ and \\ x^{2} + 3 x y = - 18 \end{matrix}

$\begin{array}{c} x = - 2 y \\ \text{and} \\ x^{2} + 3 x y= -18 \end{array}$

First, we can substitute $- 2 y$ in place of $x$ in the second equation:

\begin{aligned} (- 2 y)^{2} + 3 (- 2 y) (y) & = - 18 \end{aligned}

$\begin{aligned} (-2y)^{2} + 3 (-2y) (y)& = -18 \end{aligned}$

We can simplify this equation to solve for $y$ :

\begin{aligned} (- 2 y)^{2} + 3 (- 2 y) (y) & = - 18 \\ 4 y^{2} - 6 y^{2} & = - 18 \\ - 2 y^{2} & = - 18 \\ y^{2} & = 9 \end{aligned}

$\begin{aligned} (-2y)^{2} + 3 (-2y) (y)& = -18 \\ 4 y^{2}-6 y^{2}& = -18 \\ - 2 y^{2}& = -18 \\ y^{2}& = 9 \end{aligned}$

$y$ can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about $y^{2}$ . So we have to consider both the positive and the negative values as solutions to the equations.

\begin{matrix} y = 3 or y = - 3 \end{matrix}

$\begin{array}{c} y = 3 \text{ or } y = -3 \end{array}$

STEP: Use the

y

$y$ -values to find the

x

$x$ -values
[−1 point ⇒ 0 / 4 points left]

Each of the $y$ -values that we have found will have a corresponding value of $x$ , which can be found from the equation $x = - 2 y$ .

For $y = 3$ :

\begin{aligned} x & = - 2 y \\ = - 2 (3) \\ = - 6 \end{aligned}

$\begin{aligned} x & = -2y \\ & = -2(3) \\ & = -6 \end{aligned}$

And for $y = -3$ :

\begin{aligned} x & = - 2 y \\ = - 2 (- 3) \\ = 6 \end{aligned}

$\begin{aligned} x & = -2y \\ & = -2(-3) \\ & = 6 \end{aligned}$

So the $x$ -values are $x = -6 \text{ or } x = 6$ .

Therefore the correct coordinate pairs are $(-6; 3)$ and $(6; -3)$ .

Submit your answer as: and

Simultaneous equations with $xy$ terms

Solve for $x$ and $y$ in the following simultaneous equations:

\begin{aligned} - y & = 9 x - x y + 1 \\ y & = - x + 3 \end{aligned}

$\begin{aligned} - y & = 9 x-x y + 1\\ y & = - x + 3 \end{aligned}$

INSTRUCTION:

You should get two coordinate pairs in the form $\left(x_1; y_1\right)$ and $\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting $- x + 3$ in place of $y$ in the hyperbolic equation and then solve for $x$ .

STEP: Substitute the linear equation into the hyperbolic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

\begin{aligned} hyperbolic equation ⟶ & - y = 9 x - x y + 1 \\ linear equation ⟶ & y = - x + 3 \end{aligned}

$\begin{aligned} {\text{hyperbolic equation}}\quad \longrightarrow \quad & - y = 9 x-x y + 1\\ {\text{linear equation}}\quad \longrightarrow \quad & y = - x + 3 \end{aligned}$

This means we want two pairs of $x$ and $y$ -values which solve both equations.

We will substitute the value of $y$ from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

NOTE: We could also isolate either

x

$x$ or

y

$y$ from

- y = 9 x - x y + 1

$- y = 9 x-x y + 1$ and substitute its value into

y = - x + 3

$y = - x + 3$ . We could still get the same answer, but the calculations might become too long.

\begin{aligned} - (- x + 3) & = 9 x - x (- x + 3) + 1 \\ 0 & = 9 x - x (- x + 3) + 1 + (- x + 3) \\ 0 & = 9 x - (- x^{2} + 3 x) + 1 + (- x + 3) \\ 0 & = x^{2} + 6 x - x + 1 + 3 \\ 0 & = x^{2} + 5 x + 1 + 3 \\ 0 & = x^{2} + 5 x + 4 \end{aligned}

$\begin{aligned} - (- x + 3) & = 9 x-x (- x + 3) + 1\\ 0 & = 9 x-x (- x + 3) + 1+(- x + 3) \\ 0 & = {{{{{9}\,{}{x}}-{\left({{-{{x}^{2}}}+{{3}\,{}{x}}}\right)}}+{1}}+{\left({{-{x}}+{3}}\right)}}\\0 & = {{{{{{x}^{2}}+{{6}\,{}{x}}}-{x}}+{1}}+{3}}\\0 & = {{{{{x}^{2}}+{{5}\,{}{x}}}+{1}}+{3}}\\0 & = {{{{x}^{2}}+{{5}\,{}{x}}}+{4}}\\ \end{aligned}$

STEP: Solve the equation

0 = x^{2} + 5 x + 4

$0 = {{{{x}^{2}}+{{5}\,{}{x}}}+{4}}$
[−2 points ⇒ 2 / 6 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = x^{2} + 5 x + 4 \\ 0 & = (x + 1) (x + 4) \end{aligned}

$\begin{aligned} 0 & = {{{{x}^{2}}+{{5}\,{}{x}}}+{4}} \\ 0 & = (x+1)(x+4) \end{aligned}$

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$ -values of our solutions. The $x$ -values are $-\text{1}$ and $-\text{4}$ .

STEP: Calculate the corresponding values of

y

$y$
[−2 points ⇒ 0 / 6 points left]

By using the values of $x$ , we will calculate the corresponding values of $y$ . We will do so by substituting each $x$ -value into the linear equation.

NOTE: We can still get the same values of

y

$y$ if we substitute the values

x

$x$ into the equation

- y = 9 x - x y + 1

$- y = 9 x-x y + 1$ . If we do that, the calculations might become too long.

For $x = -1$ :

\begin{aligned} y & = - x + 3 \\ = 4 \end{aligned}

$\begin{aligned} y &= - x + 3 \\ &= 4 \end{aligned}$

For $x = -4$ :

\begin{aligned} y & = - x + 3 \\ = 7 \end{aligned}

$\begin{aligned} y &= - x + 3 \\ &= 7 \end{aligned}$

The solutions are $x = -1$ together with $y = 4$ , and $x = -4$ with $y = 7$ . Using coordinate pairs, the solutions are $\left(-1; 4\right)$ and $\left(-4; 7\right)$ .

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(-1; 4\right)$ and $\left(-4; 7\right)$ . These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(-1; 4\right)$ and $\left(-4; 7\right)$ .

Submit your answer as: and

Exercises

Points of intersection

Draw the graphs of the following equations on the same Cartesian plane. Use the graphs to find the points of intersection of the two lines.

\begin{aligned} y & = x^{2} + 4 x + 3 \\ y & = - x - 1 \end{aligned}

$\begin{aligned} y & = x^{2} + 4 x + 3 \\ y & = - x - 1 \end{aligned}$

INSTRUCTION:

You should get two coordinate pairs in the form $\left(x_1; y_1\right)$ and $\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The points of intersection are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to plot the graphs of the two equations on the same Cartesian plane.

STEP: Plot the graph of

y = x^{2} + 4 x + 3

$y = x^{2} + 4 x + 3$ on a Cartesian plane
[−2 points ⇒ 4 / 6 points left]

The $y$ -intercept is the quantity in the equation which is not multiplied by $x$ . For this equation, the $y$ -intercept is $\text{3}$ . The coordinates of this point are $\left(0; 3\right)$ .

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = x^{2} + 4 x + 3 \\ 0 & = x^{2} + x + 3 x + 3 \end{aligned}

$\begin{aligned} 0 &=x^{2} + 4 x + 3 \\ 0 &= x^{2}+x+3 x+3 \end{aligned}$

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

\begin{aligned} 0 & = x (x + 1) + 3 (x + 1) \\ 0 & = (x + 3) (x + 1) \end{aligned}

$\begin{aligned} 0 &= x(x+1)+3(x+1)\\ 0 &= (x+3)(x+1) \end{aligned}$

Using these $x$ - and $y$ -intercepts we will draw the graph:

STEP: Plot the graph of

y = - x - 1

$y = - x - 1$ on the same Cartesian plane
[−2 points ⇒ 2 / 6 points left]

The $y$ -intercept is $-\text{1}$ . Its coordinates are: $\left(0; -1\right)$ .

We will work out the root of the equation $0 = - x - 1$ .

\begin{aligned} 0 & = - x - 1 & = 0 \\ - x & = 1 \\ x & = - 1 \end{aligned}

$\begin{aligned} 0 & = - x - 1 &= 0\\ - x &= 1\\x &= -1 \end{aligned}$

The root is: $x = -1$ . It's coordinates are: $\left( -1; 0 \right)$ .

We will now plot the equation $y = - x - 1$ on the same Cartesian plane we have used for the graph of the quadratic equation.

STEP: Read off the coordinates of the points of intersection
[−2 points ⇒ 0 / 6 points left]

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: $(-1; 0)$ and $(-4; 3)$ .

We have used a graphical method to determine the solutions of the simultaneous equations: $y = x^{2} + 4 x + 3$ and $y = - x - 1$ . The solutions are the points where their graphs intersect.

NOTE: It is possible to solve the simultaneous equations using either the substitution or elimination methods.

The points of intersection are $(-1; 0)$ and $(-4; 3)$ .

Submit your answer as: and

Points of intersection

Sketch the graphs of the following equations on the same Cartesian plane. Use the graphs to find the points of intersection of the two lines.

\begin{aligned} y & = x^{2} + 6 x + 5 \\ y & = 2 x + 2 \end{aligned}

$\begin{aligned} y & = x^{2} + 6 x + 5 \\ y & = 2 x + 2 \end{aligned}$

INSTRUCTION:

You should get two coordinate pairs in the form $\left(x_1; y_1\right)$ and $\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The points of intersection are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to plot the graphs of the two equations on the same Cartesian plane.

STEP: Plot the graph of

y = x^{2} + 6 x + 5

$y = x^{2} + 6 x + 5$ on a Cartesian plane
[−2 points ⇒ 4 / 6 points left]

The $y$ -intercept is the quantity in the equation which is not multiplied by $x$ . For this equation, the $y$ -intercept is $\text{5}$ . The coordinates of this point are $\left(0; 5\right)$ .

The $x$ -intercepts are the roots of the equation $x^{2} + 6 x + 5 = 0$ . We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{6}$ . This is the coefficient of the middle term. The product of the two numbers must be $\text{5}$ . This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{5}$ . The two numbers are $\text{1}$ and $\text{5}$ . We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = x^{2} + 6 x + 5 \\ 0 & = x^{2} + x + 5 x + 5 \end{aligned}

$\begin{aligned} 0 &=x^{2} + 6 x + 5 \\ 0 &= x^{2}+x+5 x+5 \end{aligned}$

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

\begin{aligned} 0 & = x (x + 1) + 5 (x + 1) \\ 0 & = (x + 5) (x + 1) \end{aligned}

$\begin{aligned} 0 &= x(x+1)+5(x+1)\\ 0 &= (x+5)(x+1) \end{aligned}$

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are $x = -1$ and $x = -5$ . The coordinates of these roots are $\left(-1; 0 \right)$ and $\left(-5; 0 \right)$ .

Using these $x$ - and $y$ -intercepts we will draw the graph:

STEP: Plot the graph of

y = 2 x + 2

$y = 2 x + 2$ on the same Cartesian plane
[−2 points ⇒ 2 / 6 points left]

The $y$ -intercept is $\text{2}$ . Its coordinates are: $\left(0; 2\right)$ .

We will work out the root of the equation $0 = 2 x + 2$ .

\begin{aligned} 0 & = 2 x + 2 & = 0 \\ 2 x & = - 2 \\ x & = - 1 \end{aligned}

$\begin{aligned} 0 & = 2 x + 2 &= 0\\ 2 x &= -2\\x &= -1 \end{aligned}$

The root is: $x = -1$ . It's coordinates are: $\left( -1; 0 \right)$ .

We will now plot the equation $y = 2 x + 2$ on the same Cartesian plane we have used for the graph of the quadratic equation.

STEP: Read off the coordinates of the points of intersection
[−2 points ⇒ 0 / 6 points left]

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: $(-1; 0)$ and $(-3; -4)$ .

We have used a graphical method to determine the solutions of the simultaneous equations: $y = x^{2} + 6 x + 5$ and $y = 2 x + 2$ . The solutions are the points where their graphs intersect.

NOTE: It is possible to solve the simultaneous equations using either the substitution or elimination methods.

The points of intersection are $(-1; 0)$ and $(-3; -4)$ .

Submit your answer as: and

Points of intersection

Draw the graphs of the following equations on the same Cartesian plane. Use the graphs to find the points of intersection of the two lines.

\begin{aligned} y & = x^{2} + 4 x + 3 \\ y & = x + 7 \end{aligned}

$\begin{aligned} y & = x^{2} + 4 x + 3 \\ y & = x + 7 \end{aligned}$

INSTRUCTION:

You should get two coordinate pairs in the form $\left(x_1; y_1\right)$ and $\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The points of intersection are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to plot the graphs of the two equations on the same Cartesian plane.

STEP: Plot the graph of

y = x^{2} + 4 x + 3

$y = x^{2} + 4 x + 3$ on a Cartesian plane
[−2 points ⇒ 4 / 6 points left]

The $y$ -intercept is the quantity in the equation which is not multiplied by $x$ . For this equation, the $y$ -intercept is $\text{3}$ . The coordinates of this point are $\left(0; 3\right)$ .

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = x^{2} + 4 x + 3 \\ 0 & = x^{2} + x + 3 x + 3 \end{aligned}

$\begin{aligned} 0 &=x^{2} + 4 x + 3 \\ 0 &= x^{2}+x+3 x+3 \end{aligned}$

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

\begin{aligned} 0 & = x (x + 1) + 3 (x + 1) \\ 0 & = (x + 3) (x + 1) \end{aligned}

$\begin{aligned} 0 &= x(x+1)+3(x+1)\\ 0 &= (x+3)(x+1) \end{aligned}$

Using these $x$ - and $y$ -intercepts we will draw the graph:

STEP: Plot the graph of

y = x + 7

$y = x + 7$ on the same Cartesian plane
[−2 points ⇒ 2 / 6 points left]

The $y$ -intercept is $\text{7}$ . Its coordinates are: $\left(0; 7\right)$ .

We will work out the root of the equation $0 = x + 7$ .

\begin{aligned} 0 & = x + 7 & = 0 \\ x & = - 7 \end{aligned}

$\begin{aligned} 0 & = x + 7 &= 0\\ x &= -7 \end{aligned}$

The root is: $x = -7$ . It's coordinates are: $\left( -7; 0 \right)$ .

We will now plot the equation $y = x + 7$ on the same Cartesian plane we have used for the graph of the quadratic equation.

STEP: Read off the coordinates of the points of intersection
[−2 points ⇒ 0 / 6 points left]

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: $(1; 8)$ and $(-4; 3)$ .

We have used a graphical method to determine the solutions of the simultaneous equations: $y = x^{2} + 4 x + 3$ and $y = x + 7$ . The solutions are the points where their graphs intersect.

NOTE: It is possible to solve the simultaneous equations using either the substitution or elimination methods.

The points of intersection are $(1; 8)$ and $(-4; 3)$ .

Submit your answer as: and

Simultaneous equations: solving by substitution

Solve this pair of simultaneous equations. You should get two coordinate pairs, $(x;y)$ , for your answers.

\begin{aligned} x - 5 & = - 3 y - 2 \\ x + 5 y & = - y^{2} + 6 \end{aligned}

$\begin{aligned} x - 5 & = - 3 y - 2 \\ x + 5 y & = - y^{2} + 6 \end{aligned}$

INSTRUCTION: When you type your answers they should look something like this: ( 2; -4 ) and ( -1; 3). It does not matter which coordinate pair you type first.

Answer: The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.

STEP: Rearrange the linear equation to isolate a variable
[−1 point ⇒ 5 / 6 points left]

\begin{aligned} x - 5 & = - 3 y - 2 \\ x & = - 3 y + 3 \end{aligned}

$\begin{aligned} x - 5 & = - 3 y - 2 \\ x & = - 3 y + 3 \end{aligned}$

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 6 points left]

Now we need to substitute the entire right side of the equation, $(- 3 y + 3)$ , into the quadratic equation in place of $x$ . Be sure to use brackets! Then simplify the equation to get standard form.

\begin{aligned} x + 5 y & = - y^{2} + 6 \\ (- 3 y + 3) + 5 y & = - y^{2} + 6 \\ 2 y + 3 & = - y^{2} + 6 \\ y^{2} + 2 y - 3 & = 0 \leftarrow_{to keep the quadratic term positive}^{move the terms onto the left side} \end{aligned}

$\begin{aligned} x + 5 y & = - y^{2} + 6 \\ \left(- 3 y + 3\right) + 5 y & = - y^{2} + 6 \\ 2 y + 3 & = - y^{2} + 6 \\ y^{2} + 2 y - 3 & = 0 \quad \leftarrow \;\; ^{\text{move the terms onto the left side}}_{\text{to keep the quadratic term positive}} \end{aligned}$

STEP: Solve for the values of

y

$y$
[−1 point ⇒ 2 / 6 points left]

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two $y$ -values.

\begin{aligned} y^{2} + 2 y - 3 & = 0 \\ (y - 1) (y + 3) & = 0 \\ y = 1 & and y = - 3 \end{aligned}

$\begin{aligned} y^{2} + 2 y - 3 & = 0 \\ \left(y - 1\right) \left(y + 3\right) & = 0 \\ y = 1 \quad & \text{ and } \quad y = -3 \end{aligned}$

STEP: Find the value of

x

$x$ for each of the

y

$y$ -values
[−2 points ⇒ 0 / 6 points left]

Cool! However, there is still work to do. We must now find the values of $x$ for these equations. We must substitute in the values $y = 1$ and $y = -3$ . We can use either equation to do this, so we will choose the easier option: the linear equation.

\begin{aligned} If y = 1 : \\ x & = - 3 (1) + 3 \\ x & = 0 \\ If y = - 3 : \\ x & = - 3 (- 3) + 3 \\ x & = 12 \end{aligned}

$\begin{aligned} \text{If } \; y = 1: \; & \\ x & = - 3 (1) + 3 \\ x & = 0 \\ \text{If } \; y = -3: \; & \\ x & = - 3 (-3) + 3 \\ x & = 12 \end{aligned}$

Finally we have the answers: if $x = 0$ then $y = 1$ and if $x = 12$ then $y = -3$ .

NOTE: The graph below is not required for the solution. It is here to show the connection between the answers and the equations.

The solutions are: $\left(0 ; 1\right)$ and $\left(12 ; -3\right)$ .

Submit your answer as: and

Simultaneous equations: solving by substitution

Solve simultaneously for $x$ and $y$ . You should get two coordinate pairs, $(x;y)$ , for your answers.

\begin{aligned} - 3 x + y + 16 & = 0 \\ 0 & = x^{2} - y - 16 \end{aligned}

$\begin{aligned} - 3 x + y + 16 & = 0 \\ 0 & = x^{2} - y - 16 \end{aligned}$

INSTRUCTION: When you type your answers they should look something like this: ( 2; -4 ) and ( -1; 3). It does not matter which coordinate pair you type first.

Answer: The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.

STEP: Rearrange the linear equation to isolate a variable
[−1 point ⇒ 5 / 6 points left]

\begin{aligned} - 3 x + y + 16 & = 0 \\ y & = 3 x - 16 \end{aligned}

$\begin{aligned} - 3 x + y + 16 & = 0 \\ y & = 3 x - 16 \end{aligned}$

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 6 points left]

Now we need to substitute the entire right side of the equation, $(3 x - 16)$ , into the quadratic equation in place of $y$ . Be sure to use brackets! Then simplify the equation to get standard form.

\begin{aligned} 0 & = x^{2} - y - 16 \\ 0 & = x^{2} - (3 x - 16) - 16 \\ 0 & = x^{2} - 3 x \end{aligned}

$\begin{aligned} 0 & = x^{2} - y - 16 \\ 0 & = x^{2} - \left(3 x - 16\right) - 16 \\ 0 & = x^{2} - 3 x \end{aligned}$

STEP: Solve for the values of

x

$x$
[−1 point ⇒ 2 / 6 points left]

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two $x$ -values.

\begin{aligned} 0 & = x^{2} - 3 x \\ 0 & = x (x - 3) \\ x = 3 & and x = 0 \end{aligned}

$\begin{aligned} 0 & = x^{2} - 3 x \\ 0 & = x \left(x - 3\right) \\ x = 3 \quad & \text{ and } \quad x = 0 \end{aligned}$

STEP: Find the value of

y

$y$ for each of the

x

$x$ -values
[−2 points ⇒ 0 / 6 points left]

Cool! However, there is still work to do. We must now find the values of $y$ for these equations. We must substitute in the values $x = 3$ and $x = 0$ . We can use either equation to do this, so we will choose the easier option: the linear equation.

\begin{aligned} If x = 3 : \\ y & = 3 (3) - 16 \\ y & = - 7 \\ If x = 0 : \\ y & = 3 (0) - 16 \\ y & = - 16 \end{aligned}

$\begin{aligned} \text{If } \; x = 3: \; & \\ y & = 3 (3) - 16 \\ y & = -7 \\ \text{If } \; x = 0: \; & \\ y & = 3 (0) - 16 \\ y & = -16 \end{aligned}$

Finally we have the answers: if $x = 3$ then $y = -7$ and if $x = 0$ then $y = -16$ .

NOTE: The graph below is not required for the solution. It is here to show the connection between the answers and the equations.

The solutions are: $\left(3 ; -7\right)$ and $\left(0 ; -16\right)$ .

Submit your answer as: and

Simultaneous equations: solving by substitution

Find the complete solution for these simultaneous equations. You should get two coordinate pairs, $(x;y)$ , for your answers.

\begin{aligned} 0 & = - x - y + 20 \\ y^{2} - 8 & = - x + 6 y \end{aligned}

$\begin{aligned} 0 & = - x - y + 20 \\ y^{2} - 8 & = - x + 6 y \end{aligned}$

INSTRUCTION: When you type your answers they should look something like this: ( 2; -4 ) and ( -1; 3). It does not matter which coordinate pair you type first.

Answer: The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.

STEP: Rearrange the linear equation to isolate a variable
[−1 point ⇒ 5 / 6 points left]

\begin{aligned} 0 & = - x - y + 20 \\ x & = - y + 20 \end{aligned}

$\begin{aligned} 0 & = - x - y + 20 \\ x & = - y + 20 \end{aligned}$

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 6 points left]

Now we need to substitute the entire right side of the equation, $(- y + 20)$ , into the quadratic equation in place of $x$ . Be sure to use brackets! Then simplify the equation to get standard form.

\begin{aligned} y^{2} - 8 & = - x + 6 y \\ y^{2} - 8 & = - (- y + 20) + 6 y \\ y^{2} - 8 & = 7 y - 20 \\ y^{2} - 7 y + 12 & = 0 \leftarrow_{to keep the quadratic term positive}^{move the terms onto the left side} \end{aligned}

$\begin{aligned} y^{2} - 8 & = - x + 6 y \\ y^{2} - 8 & = - \left(- y + 20\right) + 6 y \\ y^{2} - 8 & = 7 y - 20 \\ y^{2} - 7 y + 12 & = 0 \quad \leftarrow \;\; ^{\text{move the terms onto the left side}}_{\text{to keep the quadratic term positive}} \end{aligned}$

STEP: Solve for the values of

y

$y$
[−1 point ⇒ 2 / 6 points left]

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two $y$ -values.

\begin{aligned} y^{2} - 7 y + 12 & = 0 \\ (y - 4) (y - 3) & = 0 \\ y = 4 & and y = 3 \end{aligned}

$\begin{aligned} y^{2} - 7 y + 12 & = 0 \\ \left(y - 4\right) \left(y - 3\right) & = 0 \\ y = 4 \quad & \text{ and } \quad y = 3 \end{aligned}$

STEP: Find the value of

x

$x$ for each of the

y

$y$ -values
[−2 points ⇒ 0 / 6 points left]

Cool! However, there is still work to do. We must now find the values of $x$ for these equations. We must substitute in the values $y = 4$ and $y = 3$ . We can use either equation to do this, so we will choose the easier option: the linear equation.

\begin{aligned} If y = 4 : \\ x & = - (4) + 20 \\ x & = 16 \\ If y = 3 : \\ x & = - (3) + 20 \\ x & = 17 \end{aligned}

$\begin{aligned} \text{If } \; y = 4: \; & \\ x & = - (4) + 20 \\ x & = 16 \\ \text{If } \; y = 3: \; & \\ x & = - (3) + 20 \\ x & = 17 \end{aligned}$

Finally we have the answers: if $x = 16$ then $y = 4$ and if $x = 17$ then $y = 3$ .

NOTE: The graph below is not required for the solution. It is here to show the connection between the answers and the equations.

The solutions are: $\left(16 ; 4\right)$ and $\left(17 ; 3\right)$ .

Submit your answer as: and

Simultaneous equations

Solve for $x$ and $y$ in the following simultaneous equations:

\begin{aligned} y & = x^{2} - 4 x \\ y & = x - 4 \end{aligned}

$\begin{aligned} y & = x^{2} - 4 x \\ y & = x - 4 \end{aligned}$

INSTRUCTION: You should get two coordinate pairs in the form

(x_{1}; y_{1})

$\left(x_1; y_1\right)$ and

(x_{2}; y_{2})

$\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided. It does not matter which coordinate pair you type first.

Answer:

The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting $x - 4$ in place of $y$ in the quadratic equation and then simplify.

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

\begin{aligned} quadratic equation ⟶ & y = x^{2} - 4 x \\ linear equation ⟶ & y = x - 4 \end{aligned}

$\begin{aligned} {\text{quadratic equation}} \quad \longrightarrow \quad & y = x^{2} - 4 x \\ {\text{linear equation}} \quad \longrightarrow \quad & y = x - 4 \end{aligned}$

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

\begin{aligned} x - 4 & = x^{2} - 4 x \\ 0 & = x^{2} - 4 x - x + 4 \\ 0 & = x^{2} - 4 x - x + 4 \\ 0 & = x^{2} - 5 x + 4 \end{aligned}

$\begin{aligned} x - 4 &= x^{2} - 4 x \\ 0 &= x^{2} - 4 x-x+4 \\ 0 &= x^{2}-4 x-x+4\\ 0 &= x^{2}-5 x+4 \end{aligned}$

STEP: Solve the equation

x^{2} - 5 x + 4 = 0

$x^{2}-5 x+4 = 0$
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{5}$ . This is the coefficient of the middle term. The product of the two numbers must be $\text{4}$ . This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{4}$ . The two numbers are $-\text{1}$ and $-\text{4}$ . We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = x^{2} - 5 x + 4 \\ 0 & = x^{2} - x - 4 x + 4 \end{aligned}

$\begin{aligned} 0 &= x^{2}-5 x+4 \\ 0 &= x^{2}-x-4 x+4 \end{aligned}$

We now group the first two terms together and group the last terms together and factorise.

\begin{aligned} 0 & = x^{2} - x - 4 x + 4 \\ 0 & = x (x - 1) - 4 (x - 1) \end{aligned}

$\begin{aligned} 0 &= \color{red}{x^{2}-x}\color{blue}{-4 x+4}\\ 0 &= \color{red}{x(x-1)}\color{blue}{-4(x-1)} \end{aligned}$

The last thing now is to factor the highest common factor.

\begin{aligned} 0 & = (x - 4) (x - 1) \end{aligned}

$\begin{aligned} 0 &= (x-4)(x-1) \end{aligned}$

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$ -values of our solutions. The $x$ -values are $\text{4}$ and $\text{1}$ .

STEP: Calculate the corresponding values of

y

$y$
[−2 points ⇒ 0 / 6 points left]

We will substitute each $x$ -value into the linear equation. This will give us the corresponding values of $y$ .

For $x = 4$ :

\begin{aligned} y & = x - 4 \\ = 0 \end{aligned}

$\begin{aligned} y &= x - 4 \\ &= 0 \end{aligned}$

For $x = 1$ :

\begin{aligned} y & = x - 4 \\ = - 3 \end{aligned}

$\begin{aligned} y &= x - 4 \\ &= -3 \end{aligned}$

The solutions are $x = 4$ together with $y = 0$ , and $x = 1$ with $y = -3$ . In coordinates pairs, the solutions are $\left(4; 0\right)$ and $\left(1; -3\right)$ .

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(4; 0\right)$ and $\left(1; -3\right)$ . These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(4; 0\right)$ and $\left(1; -3\right)$ .

Submit your answer as: and

Simultaneous equations

Solve the following equations simultaneously:

\begin{aligned} y & = x^{2} + 5 x + 4 \\ y & = x + 1 \end{aligned}

$\begin{aligned} y & = x^{2} + 5 x + 4 \\ y & = x + 1 \end{aligned}$

INSTRUCTION: You should get two coordinate pairs in the form

(x_{1}; y_{1})

$\left(x_1; y_1\right)$ and

(x_{2}; y_{2})

$\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided. It does not matter which coordinate pair you type first.

Answer:

The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting $x + 1$ in place of $y$ in the quadratic equation and then simplify.

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

\begin{aligned} quadratic equation ⟶ & y = x^{2} + 5 x + 4 \\ linear equation ⟶ & y = x + 1 \end{aligned}

$\begin{aligned} {\text{quadratic equation}} \quad \longrightarrow \quad & y = x^{2} + 5 x + 4 \\ {\text{linear equation}} \quad \longrightarrow \quad & y = x + 1 \end{aligned}$

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

\begin{aligned} x + 1 & = x^{2} + 5 x + 4 \\ 0 & = x^{2} + 5 x + 4 - x - 1 \\ 0 & = x^{2} + 5 x - x + 4 - 1 \\ 0 & = x^{2} + 4 x + 3 \end{aligned}

$\begin{aligned} x + 1 &= x^{2} + 5 x + 4 \\ 0 &= x^{2} + 5 x + 4-x-1 \\ 0 &= x^{2}+5 x-x+4-1\\ 0 &= x^{2}+4 x+3 \end{aligned}$

STEP: Solve the equation

x^{2} + 4 x + 3 = 0

$x^{2}+4 x+3 = 0$
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{4}$ . This is the coefficient of the middle term. The product of the two numbers must be $\text{3}$ . This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{3}$ . The two numbers are $\text{1}$ and $\text{3}$ . We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = x^{2} + 4 x + 3 \\ 0 & = x^{2} + x + 3 x + 3 \end{aligned}

$\begin{aligned} 0 &= x^{2}+4 x+3 \\ 0 &= x^{2}+x+3 x+3 \end{aligned}$

We now group the first two terms together and group the last terms together and factorise.

\begin{aligned} 0 & = x^{2} + x + 3 x + 3 \\ 0 & = x (x + 1) + 3 (x + 1) \end{aligned}

$\begin{aligned} 0 &= \color{red}{x^{2}+x}\color{blue}{+3 x+3}\\ 0 &= \color{red}{x(x+1)}\color{blue}{+3(x+1)} \end{aligned}$

The last thing now is to factor the highest common factor.

\begin{aligned} 0 & = (x + 3) (x + 1) \end{aligned}

$\begin{aligned} 0 &= (x+3)(x+1) \end{aligned}$

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$ -values of our solutions. The $x$ -values are $-\text{1}$ and $-\text{3}$ .

STEP: Calculate the corresponding values of

y

$y$
[−2 points ⇒ 0 / 6 points left]

We will substitute each $x$ -value into the linear equation. This will give us the corresponding values of $y$ .

For $x = -1$ :

\begin{aligned} y & = x + 1 \\ = 0 \end{aligned}

$\begin{aligned} y &= x + 1 \\ &= 0 \end{aligned}$

For $x = -3$ :

\begin{aligned} y & = x + 1 \\ = - 2 \end{aligned}

$\begin{aligned} y &= x + 1 \\ &= -2 \end{aligned}$

The solutions are $x = -1$ together with $y = 0$ , and $x = -3$ with $y = -2$ . In coordinates pairs, the solutions are $\left(-1; 0\right)$ and $\left(-3; -2\right)$ .

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(-1; 0\right)$ and $\left(-3; -2\right)$ . These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(-1; 0\right)$ and $\left(-3; -2\right)$ .

Submit your answer as: and

Simultaneous equations

Solve for $x$ and $y$ from the given equations:

\begin{aligned} y & = 3 x^{2} - 10 x + 6 \\ y & = 2 x - 3 \end{aligned}

$\begin{aligned} y & = 3 x^{2} - 10 x + 6 \\ y & = 2 x - 3 \end{aligned}$

INSTRUCTION: You should get two coordinate pairs in the form

(x_{1}; y_{1})

$\left(x_1; y_1\right)$ and

(x_{2}; y_{2})

$\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided. It does not matter which coordinate pair you type first.

Answer:

The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting $2 x - 3$ in place of $y$ in the quadratic equation and then simplify.

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

\begin{aligned} quadratic equation ⟶ & y = 3 x^{2} - 10 x + 6 \\ linear equation ⟶ & y = 2 x - 3 \end{aligned}

$\begin{aligned} {\text{quadratic equation}} \quad \longrightarrow \quad & y = 3 x^{2} - 10 x + 6 \\ {\text{linear equation}} \quad \longrightarrow \quad & y = 2 x - 3 \end{aligned}$

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

\begin{aligned} 2 x - 3 & = 3 x^{2} - 10 x + 6 \\ 0 & = 3 x^{2} - 10 x + 6 - 2 x + 3 \\ 0 & = 3 x^{2} - 10 x - 2 x + 6 + 3 \\ 0 & = 3 x^{2} - 12 x + 9 \end{aligned}

$\begin{aligned} 2 x - 3 &= 3 x^{2} - 10 x + 6 \\ 0 &= 3 x^{2} - 10 x + 6-2 x+3 \\ 0 &= 3 x^{2}-10 x-2 x+6+3\\ 0 &= 3 x^{2}-12 x+9 \end{aligned}$

STEP: Solve the equation

3 x^{2} - 12 x + 9 = 0

$3 x^{2}-12 x+9 = 0$
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{12}$ . This is the coefficient of the middle term. The product of the two numbers must be $\text{27}$ . This is the product of the coefficient of the first term $\text{3}$ and the value of the last term $\text{9}$ . The two numbers are $-\text{3}$ and $-\text{9}$ . We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = 3 x^{2} - 12 x + 9 \\ 0 & = 3 x^{2} - 3 x - 9 x + 9 \end{aligned}

$\begin{aligned} 0 &= 3 x^{2}-12 x+9 \\ 0 &= 3 x^{2}-3 x-9 x+9 \end{aligned}$

We now group the first two terms together and group the last terms together and factorise.

\begin{aligned} 0 & = 3 x^{2} - 3 x - 9 x + 9 \\ 0 & = 3 x (x - 1) - 9 (x - 1) \end{aligned}

$\begin{aligned} 0 &= \color{red}{3 x^{2}-3 x}\color{blue}{-9 x+9}\\ 0 &= \color{red}{3 x(x-1)}\color{blue}{-9(x-1)} \end{aligned}$

The last thing now is to factor the highest common factor.

\begin{aligned} 0 & = (3 x - 9) (x - 1) \end{aligned}

$\begin{aligned} 0 &= (3 x-9)(x-1) \end{aligned}$

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$ -values of our solutions. The $x$ -values are $\text{3}$ and $\text{1}$ .

STEP: Calculate the corresponding values of

y

$y$
[−2 points ⇒ 0 / 6 points left]

We will substitute each $x$ -value into the linear equation. This will give us the corresponding values of $y$ .

For $x = 3$ :

\begin{aligned} y & = 2 x - 3 \\ = 3 \end{aligned}

$\begin{aligned} y &= 2 x - 3 \\ &= 3 \end{aligned}$

For $x = 1$ :

\begin{aligned} y & = 2 x - 3 \\ = - 1 \end{aligned}

$\begin{aligned} y &= 2 x - 3 \\ &= -1 \end{aligned}$

The solutions are $x = 3$ together with $y = 3$ , and $x = 1$ with $y = -1$ . In coordinates pairs, the solutions are $\left(3; 3\right)$ and $\left(1; -1\right)$ .

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(3; 3\right)$ and $\left(1; -1\right)$ . These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(3; 3\right)$ and $\left(1; -1\right)$ .

Submit your answer as: and

Simultaneous equations

Solve for $x$ and $y$ from the given equations:

\begin{aligned} y & = 2 x^{2} - 2 x - 1 \\ y & = - x + 2 \end{aligned}

$\begin{aligned} y & = 2 x^{2} - 2 x - 1 \\ y & = - x + 2 \end{aligned}$

INSTRUCTION:

You should get two coordinate pairs in the form $\left(x_1; y_1\right)$ and $\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting $- x + 2$ in place of $y$ in the quadratic equation and then simplify.

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

\begin{aligned} quadratic equation ⟶ & y = 2 x^{2} - 2 x - 1 \\ linear equation ⟶ & y = - x + 2 \end{aligned}

$\begin{aligned} {\text{quadratic equation}} \quad \longrightarrow \quad & y = 2 x^{2} - 2 x - 1 \\ {\text{linear equation}} \quad \longrightarrow \quad & y = - x + 2 \end{aligned}$

This means we want two pairs of $x$ and $y$ -values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of

y

$y$ from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.

\begin{aligned} - x + 2 & = 2 x^{2} - 2 x - 1 \\ 0 & = 2 x^{2} - 2 x - 1 + x - 2 \\ 0 & = 2 x^{2} - 2 x + x - 1 - 2 \\ 0 & = 2 x^{2} - x - 3 \end{aligned}

$\begin{aligned} - x + 2 &= 2 x^{2} - 2 x - 1 \\ 0 &= 2 x^{2} - 2 x - 1+x-2 \\ 0 &= 2 x^{2}-2 x+x-1-2\\ 0 &= 2 x^{2}-x-3 \end{aligned}$

STEP: Solve the equation

0 = 2 x^{2} - x - 3

$0 = 2 x^{2}-x-3$
[−2 points ⇒ 2 / 6 points left]

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{1}$ . This is the coefficient of the middle term. The product of the two numbers must be $-\text{6}$ . This is the product of the coefficient of the first term $\text{2}$ and the value of the last term $-\text{3}$ . The two numbers are $\text{2}$ and $-\text{3}$ . We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = 2 x^{2} - x - 3 \\ 0 & = 2 x^{2} + 2 x - 3 x - 3 \end{aligned}

$\begin{aligned} 0 &= 2 x^{2}-x-3 \\ 0 &= 2 x^{2}+2 x-3 x-3 \end{aligned}$

We now group the first two terms together and group the last terms together and factorise.

\begin{aligned} 0 & = 2 x^{2} + 2 x - 3 x - 3 \\ 0 & = 2 x (x + 1) - 3 (x + 1) \end{aligned}

$\begin{aligned} 0 &= \color{red}{2 x^{2}+2 x}\color{blue}{-3 x-3}\\ 0 &= \color{red}{2 x(x+1)}\color{blue}{-3(x+1)} \end{aligned}$

The last thing now is to factor the highest common factor.

\begin{aligned} 0 & = (2 x - 3) (x + 1) \end{aligned}

$\begin{aligned} 0 &= (2 x-3)(x+1) \end{aligned}$

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$ -values of our solutions. The $x$ -values are $\dfrac{3}{2}$ and $-\text{1}$ .

STEP: Calculate the corresponding values of

y

$y$
[−2 points ⇒ 0 / 6 points left]

We will substitute each $x$ -value into the linear equation. This will give us the corresponding values of $y$ .

NOTE: We can do that by substituting

x

$x$ into any of the original equations. It is easier to use the linear one.

For $x = \dfrac{3}{2}$ :

\begin{aligned} y & = - x + 2 \\ = \frac{1}{2} \end{aligned}

$\begin{aligned} y &= - x + 2 \\ &= \dfrac{1}{2} \end{aligned}$

For $x = -1$ :

\begin{aligned} y & = - x + 2 \\ = 3 \end{aligned}

$\begin{aligned} y &= - x + 2 \\ &= 3 \end{aligned}$

The solutions are $x = \dfrac{3}{2}$ together with $y = \dfrac{1}{2}$ , and $x = -1$ with $y = 3$ . In coordinate pairs, the solutions are $\left(\dfrac{3}{2}; \dfrac{1}{2}\right)$ and $\left(-1; 3\right)$ .

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(\dfrac{3}{2}; \dfrac{1}{2}\right)$ and $\left(-1; 3\right)$ . These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(\dfrac{3}{2}; \dfrac{1}{2}\right)$ and $\left(-1; 3\right)$ .

Submit your answer as: and

Simultaneous equations

Solve for $x$ and $y$ in the following simultaneous equations:

\begin{aligned} y & = 2 x^{2} - 4 x - 2 \\ y & = - x + 3 \end{aligned}

$\begin{aligned} y & = 2 x^{2} - 4 x - 2 \\ y & = - x + 3 \end{aligned}$

INSTRUCTION:

You should get two coordinate pairs in the form $\left(x_1; y_1\right)$ and $\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting $- x + 3$ in place of $y$ in the quadratic equation and then simplify.

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

\begin{aligned} quadratic equation ⟶ & y = 2 x^{2} - 4 x - 2 \\ linear equation ⟶ & y = - x + 3 \end{aligned}

$\begin{aligned} {\text{quadratic equation}} \quad \longrightarrow \quad & y = 2 x^{2} - 4 x - 2 \\ {\text{linear equation}} \quad \longrightarrow \quad & y = - x + 3 \end{aligned}$

This means we want two pairs of $x$ and $y$ -values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of

y

$y$ from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.

\begin{aligned} - x + 3 & = 2 x^{2} - 4 x - 2 \\ 0 & = 2 x^{2} - 4 x - 2 + x - 3 \\ 0 & = 2 x^{2} - 4 x + x - 2 - 3 \\ 0 & = 2 x^{2} - 3 x - 5 \end{aligned}

$\begin{aligned} - x + 3 &= 2 x^{2} - 4 x - 2 \\ 0 &= 2 x^{2} - 4 x - 2+x-3 \\ 0 &= 2 x^{2}-4 x+x-2-3\\ 0 &= 2 x^{2}-3 x-5 \end{aligned}$

STEP: Solve the equation

0 = 2 x^{2} - 3 x - 5

$0 = 2 x^{2}-3 x-5$
[−2 points ⇒ 2 / 6 points left]

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{3}$ . This is the coefficient of the middle term. The product of the two numbers must be $-\text{10}$ . This is the product of the coefficient of the first term $\text{2}$ and the value of the last term $-\text{5}$ . The two numbers are $\text{2}$ and $-\text{5}$ . We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = 2 x^{2} - 3 x - 5 \\ 0 & = 2 x^{2} + 2 x - 5 x - 5 \end{aligned}

$\begin{aligned} 0 &= 2 x^{2}-3 x-5 \\ 0 &= 2 x^{2}+2 x-5 x-5 \end{aligned}$

We now group the first two terms together and group the last terms together and factorise.

\begin{aligned} 0 & = 2 x^{2} + 2 x - 5 x - 5 \\ 0 & = 2 x (x + 1) - 5 (x + 1) \end{aligned}

$\begin{aligned} 0 &= \color{red}{2 x^{2}+2 x}\color{blue}{-5 x-5}\\ 0 &= \color{red}{2 x(x+1)}\color{blue}{-5(x+1)} \end{aligned}$

The last thing now is to factor the highest common factor.

\begin{aligned} 0 & = (2 x - 5) (x + 1) \end{aligned}

$\begin{aligned} 0 &= (2 x-5)(x+1) \end{aligned}$

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$ -values of our solutions. The $x$ -values are $\dfrac{5}{2}$ and $-\text{1}$ .

STEP: Calculate the corresponding values of

y

$y$
[−2 points ⇒ 0 / 6 points left]

We will substitute each $x$ -value into the linear equation. This will give us the corresponding values of $y$ .

NOTE: We can do that by substituting

x

$x$ into any of the original equations. It is easier to use the linear one.

For $x = \dfrac{5}{2}$ :

\begin{aligned} y & = - x + 3 \\ = \frac{1}{2} \end{aligned}

$\begin{aligned} y &= - x + 3 \\ &= \dfrac{1}{2} \end{aligned}$

For $x = -1$ :

\begin{aligned} y & = - x + 3 \\ = 4 \end{aligned}

$\begin{aligned} y &= - x + 3 \\ &= 4 \end{aligned}$

The solutions are $x = \dfrac{5}{2}$ together with $y = \dfrac{1}{2}$ , and $x = -1$ with $y = 4$ . In coordinate pairs, the solutions are $\left(\dfrac{5}{2}; \dfrac{1}{2}\right)$ and $\left(-1; 4\right)$ .

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(\dfrac{5}{2}; \dfrac{1}{2}\right)$ and $\left(-1; 4\right)$ . These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(\dfrac{5}{2}; \dfrac{1}{2}\right)$ and $\left(-1; 4\right)$ .

Submit your answer as: and

Simultaneous equations

Solve for $x$ and $y$ in the following simultaneous equations:

\begin{aligned} y & = 2 x^{2} - 8 x + 6 \\ y & = x - 1 \end{aligned}

$\begin{aligned} y & = 2 x^{2} - 8 x + 6 \\ y & = x - 1 \end{aligned}$

INSTRUCTION:

You should get two coordinate pairs in the form $\left(x_1; y_1\right)$ and $\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting $x - 1$ in place of $y$ in the quadratic equation and then simplify.

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

\begin{aligned} quadratic equation ⟶ & y = 2 x^{2} - 8 x + 6 \\ linear equation ⟶ & y = x - 1 \end{aligned}

$\begin{aligned} {\text{quadratic equation}} \quad \longrightarrow \quad & y = 2 x^{2} - 8 x + 6 \\ {\text{linear equation}} \quad \longrightarrow \quad & y = x - 1 \end{aligned}$

This means we want two pairs of $x$ and $y$ -values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of

y

$y$ from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.

\begin{aligned} x - 1 & = 2 x^{2} - 8 x + 6 \\ 0 & = 2 x^{2} - 8 x + 6 - x + 1 \\ 0 & = 2 x^{2} - 8 x - x + 6 + 1 \\ 0 & = 2 x^{2} - 9 x + 7 \end{aligned}

$\begin{aligned} x - 1 &= 2 x^{2} - 8 x + 6 \\ 0 &= 2 x^{2} - 8 x + 6-x+1 \\ 0 &= 2 x^{2}-8 x-x+6+1\\ 0 &= 2 x^{2}-9 x+7 \end{aligned}$

STEP: Solve the equation

0 = 2 x^{2} - 9 x + 7

$0 = 2 x^{2}-9 x+7$
[−2 points ⇒ 2 / 6 points left]

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{9}$ . This is the coefficient of the middle term. The product of the two numbers must be $\text{14}$ . This is the product of the coefficient of the first term $\text{2}$ and the value of the last term $\text{7}$ . The two numbers are $-\text{2}$ and $-\text{7}$ . We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = 2 x^{2} - 9 x + 7 \\ 0 & = 2 x^{2} - 2 x - 7 x + 7 \end{aligned}

$\begin{aligned} 0 &= 2 x^{2}-9 x+7 \\ 0 &= 2 x^{2}-2 x-7 x+7 \end{aligned}$

We now group the first two terms together and group the last terms together and factorise.

\begin{aligned} 0 & = 2 x^{2} - 2 x - 7 x + 7 \\ 0 & = 2 x (x - 1) - 7 (x - 1) \end{aligned}

$\begin{aligned} 0 &= \color{red}{2 x^{2}-2 x}\color{blue}{-7 x+7}\\ 0 &= \color{red}{2 x(x-1)}\color{blue}{-7(x-1)} \end{aligned}$

The last thing now is to factor the highest common factor.

\begin{aligned} 0 & = (2 x - 7) (x - 1) \end{aligned}

$\begin{aligned} 0 &= (2 x-7)(x-1) \end{aligned}$

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$ -values of our solutions. The $x$ -values are $\dfrac{7}{2}$ and $\text{1}$ .

STEP: Calculate the corresponding values of

y

$y$
[−2 points ⇒ 0 / 6 points left]

We will substitute each $x$ -value into the linear equation. This will give us the corresponding values of $y$ .

NOTE: We can do that by substituting

x

$x$ into any of the original equations. It is easier to use the linear one.

For $x = \dfrac{7}{2}$ :

\begin{aligned} y & = x - 1 \\ = \frac{5}{2} \end{aligned}

$\begin{aligned} y &= x - 1 \\ &= \dfrac{5}{2} \end{aligned}$

For $x = 1$ :

\begin{aligned} y & = x - 1 \\ = 0 \end{aligned}

$\begin{aligned} y &= x - 1 \\ &= 0 \end{aligned}$

The solutions are $x = \dfrac{7}{2}$ together with $y = \dfrac{5}{2}$ , and $x = 1$ with $y = 0$ . In coordinate pairs, the solutions are $\left(\dfrac{7}{2}; \dfrac{5}{2}\right)$ and $\left(1; 0\right)$ .

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(\dfrac{7}{2}; \dfrac{5}{2}\right)$ and $\left(1; 0\right)$ . These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(\dfrac{7}{2}; \dfrac{5}{2}\right)$ and $\left(1; 0\right)$ .

Submit your answer as: and

Simultaneous equations

Solve for $x$ and $y$ in the following simultaneous equations:

\begin{aligned} y & = 3 x^{2} - 5 x + 2 \\ x & = y - 2 \end{aligned}

$\begin{aligned} y & = 3 x^{2} - 5 x + 2 \\ x & = y - 2 \end{aligned}$

INSTRUCTION:

You should get two coordinate pairs in the form $\left(x_1; y_1\right)$ and $\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting $y - 2$ in place of $x$ in the quadratic equation and then simplify.

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

\begin{aligned} quadratic equation ⟶ & y = 3 x^{2} - 5 x + 2 \\ linear equation ⟶ & x = y - 2 \end{aligned}

$\begin{aligned} {\text{quadratic equation}} \quad \longrightarrow \quad & y = 3 x^{2} - 5 x + 2 \\ {\text{linear equation}} \quad \longrightarrow \quad & x = y - 2 \end{aligned}$

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $x$ from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of

y

$y$ from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.

\begin{aligned} y & = 3 {(y - 2)}^{2} - 5 (y - 2) + 2 \\ 0 & = (3) {(y - 2)}^{2} - 5 (y - 2) + 2 - y \\ 0 & = 3 (y^{2} - 4 y + 4) - (5 y - 10) + 2 - y \\ 0 & = 3 y^{2} - 12 y + 12 - (5 y - 10) + 2 - y \\ 0 & = 3 y^{2} - 12 y + 12 - 5 y + 10 + 2 - y \\ 0 & = 3 y^{2} - 12 y - 5 y - y + 12 + 10 + 2 \\ 0 & = 3 y^{2} - 17 y - y + 12 + 10 + 2 \\ 0 & = 3 y^{2} - 18 y + 12 + 10 + 2 \\ 0 & = 3 y^{2} - 18 y + 22 + 2 \\ 0 & = 3 y^{2} - 18 y + 24 \end{aligned}

$\begin{aligned} y &= 3\left(y - 2\right)^2-5\left(y - 2\right)+2 \\ 0 &= {{{{{\left({3}\right)}\,{}{{\left({{y}-{2}}\right)}^{2}}}-{{5}\,{}{\left({{y}-{2}}\right)}}}+{2}}-{y}}\\0 &= {{{{{3}\,{}{\left({{{{y}^{2}}-{{4}\,{}{y}}}+{4}}\right)}}-{\left({{{5}\,{}{y}}-{10}}\right)}}+{2}}-{y}}\\0 &= {{{{{{{3}\,{}{{y}^{2}}}-{{12}\,{}{y}}}+{12}}-{\left({{{5}\,{}{y}}-{10}}\right)}}+{2}}-{y}}\\0 &= {{{{{{{{3}\,{}{{y}^{2}}}-{{12}\,{}{y}}}+{12}}-{{5}\,{}{y}}}+{10}}+{2}}-{y}}\\0 &= {{{{{{{{3}\,{}{{y}^{2}}}-{{12}\,{}{y}}}-{{5}\,{}{y}}}-{y}}+{12}}+{10}}+{2}}\\0 &= {{{{{{{3}\,{}{{y}^{2}}}-{{17}\,{}{y}}}-{y}}+{12}}+{10}}+{2}}\\0 &= {{{{{{3}\,{}{{y}^{2}}}-{{18}\,{}{y}}}+{12}}+{10}}+{2}}\\0 &= {{{{{3}\,{}{{y}^{2}}}-{{18}\,{}{y}}}+{22}}+{2}}\\0 &= {{{{3}\,{}{{y}^{2}}}-{{18}\,{}{y}}}+{24}} \end{aligned}$

STEP: Solve the equation

0 = 3 y^{2} - 18 y + 24

$0 = {{{{3}\,{}{{y}^{2}}}-{{18}\,{}{y}}}+{24}}$
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{18}$ . This is the coefficient of the middle term. The product of the two numbers must be $\text{72}$ . This is the product of the coefficient of the first term $\text{3}$ and the value of the last term $\text{24}$ . The two numbers are $-\text{6}$ and $-\text{12}$ . We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = 3 y^{2} - 18 y + 24 \\ 0 & = 3 y^{2} - 6 y - 12 y + 24 \end{aligned}

$\begin{aligned} 0 &= {{{{3}\,{}{{y}^{2}}}-{{18}\,{}{y}}}+{24}} \\ 0 &= 3 y^{2}-6 y-12 y+24 \end{aligned}$

We now group the first two terms together and group the last terms together and factorise.

\begin{aligned} 0 & = 3 y^{2} - 6 y - 12 y + 24 \\ 0 & = - 3 y (- y + 2) + 12 (- y + 2) \end{aligned}

$\begin{aligned} 0 &= \color{red}{3 y^{2}-6 y}\color{blue}{-12 y+24}\\ 0 &= \color{red}{- 3 y(- y+2)}\color{blue}{+12(- y+2)} \end{aligned}$

The last thing now is to factor the highest common factor.

\begin{aligned} 0 & = (- 3 y + 12) (- y + 2) \end{aligned}

$\begin{aligned} 0 &= (- 3 y+12)(- y+2) \end{aligned}$

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $y$ -values of our solutions. The $y$ -values are 4 and 2.

STEP: Calculate the corresponding values of

x

$x$
[−2 points ⇒ 0 / 6 points left]

We will substitute each $y$ -value into the linear equation. This will give us the corresponding values of $x$ .

NOTE: We can do that by substituting

y

$y$ into any of the original equations. It is easier to use the linear one.

For $y = 4$ :

\begin{aligned} x & = y - 2 \\ = 2 \end{aligned}

$\begin{aligned} x &= y - 2 \\ &= 2 \end{aligned}$

For $y = 2$ :

\begin{aligned} x & = y - 2 \\ = 0 \end{aligned}

$\begin{aligned} x &= y - 2 \\ &= 0 \end{aligned}$

The solutions are $x = 2$ together with $y = 4$ , and $x = 0$ with $y = 2$ . In coordinates pairs, the solutions are $\left(2; 4\right)$ and $\left(0; 2\right)$ .

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(2; 4\right)$ and $\left(0; 2\right)$ . These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(2; 4\right)$ and $\left(0; 2\right)$ .

Submit your answer as: and

Simultaneous equations

Solve for $x$ and $y$ in the following simultaneous equations:

\begin{aligned} y & = 2 x^{2} - 3 x + 1 \\ x & = y - 1 \end{aligned}

$\begin{aligned} y & = 2 x^{2} - 3 x + 1 \\ x & = y - 1 \end{aligned}$

INSTRUCTION:

You should get two coordinate pairs in the form $\left(x_1; y_1\right)$ and $\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting $y - 1$ in place of $x$ in the quadratic equation and then simplify.

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

\begin{aligned} quadratic equation ⟶ & y = 2 x^{2} - 3 x + 1 \\ linear equation ⟶ & x = y - 1 \end{aligned}

$\begin{aligned} {\text{quadratic equation}} \quad \longrightarrow \quad & y = 2 x^{2} - 3 x + 1 \\ {\text{linear equation}} \quad \longrightarrow \quad & x = y - 1 \end{aligned}$

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $x$ from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of

y

$y$ from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.

\begin{aligned} y & = 2 {(y - 1)}^{2} - 3 (y - 1) + 1 \\ 0 & = (2) {(y - 1)}^{2} - 3 (y - 1) + 1 - y \\ 0 & = 2 (y^{2} - 2 y + 1) - (3 y - 3) + 1 - y \\ 0 & = 2 y^{2} - 4 y + 2 - (3 y - 3) + 1 - y \\ 0 & = 2 y^{2} - 4 y + 2 - 3 y + 3 + 1 - y \\ 0 & = 2 y^{2} - 4 y - 3 y - y + 2 + 3 + 1 \\ 0 & = 2 y^{2} - 7 y - y + 2 + 3 + 1 \\ 0 & = 2 y^{2} - 8 y + 2 + 3 + 1 \\ 0 & = 2 y^{2} - 8 y + 5 + 1 \\ 0 & = 2 y^{2} - 8 y + 6 \end{aligned}

$\begin{aligned} y &= 2\left(y - 1\right)^2-3\left(y - 1\right)+1 \\ 0 &= {{{{{\left({2}\right)}\,{}{{\left({{y}-{1}}\right)}^{2}}}-{{3}\,{}{\left({{y}-{1}}\right)}}}+{1}}-{y}}\\0 &= {{{{{2}\,{}{\left({{{{y}^{2}}-{{2}\,{}{y}}}+{1}}\right)}}-{\left({{{3}\,{}{y}}-{3}}\right)}}+{1}}-{y}}\\0 &= {{{{{{{2}\,{}{{y}^{2}}}-{{4}\,{}{y}}}+{2}}-{\left({{{3}\,{}{y}}-{3}}\right)}}+{1}}-{y}}\\0 &= {{{{{{{{2}\,{}{{y}^{2}}}-{{4}\,{}{y}}}+{2}}-{{3}\,{}{y}}}+{3}}+{1}}-{y}}\\0 &= {{{{{{{{2}\,{}{{y}^{2}}}-{{4}\,{}{y}}}-{{3}\,{}{y}}}-{y}}+{2}}+{3}}+{1}}\\0 &= {{{{{{{2}\,{}{{y}^{2}}}-{{7}\,{}{y}}}-{y}}+{2}}+{3}}+{1}}\\0 &= {{{{{{2}\,{}{{y}^{2}}}-{{8}\,{}{y}}}+{2}}+{3}}+{1}}\\0 &= {{{{{2}\,{}{{y}^{2}}}-{{8}\,{}{y}}}+{5}}+{1}}\\0 &= {{{{2}\,{}{{y}^{2}}}-{{8}\,{}{y}}}+{6}} \end{aligned}$

STEP: Solve the equation

0 = 2 y^{2} - 8 y + 6

$0 = {{{{2}\,{}{{y}^{2}}}-{{8}\,{}{y}}}+{6}}$
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{8}$ . This is the coefficient of the middle term. The product of the two numbers must be $\text{12}$ . This is the product of the coefficient of the first term $\text{2}$ and the value of the last term $\text{6}$ . The two numbers are $-\text{2}$ and $-\text{6}$ . We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = 2 y^{2} - 8 y + 6 \\ 0 & = 2 y^{2} - 2 y - 6 y + 6 \end{aligned}

$\begin{aligned} 0 &= {{{{2}\,{}{{y}^{2}}}-{{8}\,{}{y}}}+{6}} \\ 0 &= 2 y^{2}-2 y-6 y+6 \end{aligned}$

We now group the first two terms together and group the last terms together and factorise.

\begin{aligned} 0 & = 2 y^{2} - 2 y - 6 y + 6 \\ 0 & = 2 y (y - 1) - 6 (y - 1) \end{aligned}

$\begin{aligned} 0 &= \color{red}{2 y^{2}-2 y}\color{blue}{-6 y+6}\\ 0 &= \color{red}{2 y(y-1)}\color{blue}{-6(y-1)} \end{aligned}$

The last thing now is to factor the highest common factor.

\begin{aligned} 0 & = (2 y - 6) (y - 1) \end{aligned}

$\begin{aligned} 0 &= (2 y-6)(y-1) \end{aligned}$

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $y$ -values of our solutions. The $y$ -values are 3 and 1.

STEP: Calculate the corresponding values of

x

$x$
[−2 points ⇒ 0 / 6 points left]

We will substitute each $y$ -value into the linear equation. This will give us the corresponding values of $x$ .

NOTE: We can do that by substituting

y

$y$ into any of the original equations. It is easier to use the linear one.

For $y = 3$ :

\begin{aligned} x & = y - 1 \\ = 2 \end{aligned}

$\begin{aligned} x &= y - 1 \\ &= 2 \end{aligned}$

For $y = 1$ :

\begin{aligned} x & = y - 1 \\ = 0 \end{aligned}

$\begin{aligned} x &= y - 1 \\ &= 0 \end{aligned}$

The solutions are $x = 2$ together with $y = 3$ , and $x = 0$ with $y = 1$ . In coordinates pairs, the solutions are $\left(2; 3\right)$ and $\left(0; 1\right)$ .

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(2; 3\right)$ and $\left(0; 1\right)$ . These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(2; 3\right)$ and $\left(0; 1\right)$ .

Submit your answer as: and

Simultaneous equations

Solve for $x$ and $y$ in the following simultaneous equations:

\begin{aligned} y & = x^{2} + x + 2 \\ x & = y - 3 \end{aligned}

$\begin{aligned} y & = x^{2} + x + 2 \\ x & = y - 3 \end{aligned}$

INSTRUCTION:

You should get two coordinate pairs in the form $\left(x_1; y_1\right)$ and $\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting $y - 3$ in place of $x$ in the quadratic equation and then simplify.

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

\begin{aligned} quadratic equation ⟶ & y = x^{2} + x + 2 \\ linear equation ⟶ & x = y - 3 \end{aligned}

$\begin{aligned} {\text{quadratic equation}} \quad \longrightarrow \quad & y = x^{2} + x + 2 \\ {\text{linear equation}} \quad \longrightarrow \quad & x = y - 3 \end{aligned}$

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $x$ from the linear equation into the quadratic equation. Then we will simplify.

NOTE: We could also substitute the value of

y

$y$ from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.

\begin{aligned} y & = {(y - 3)}^{2} + (y - 3) + 2 \\ 0 & = {(y - 3)}^{2} + (y - 3) + 2 - y \\ 0 & = y^{2} - 6 y + 9 + (y - 3) + 2 - y \\ 0 & = y^{2} - 6 y + 9 + y - 3 + 2 - y \\ 0 & = y^{2} - 6 y + y - y + 9 - 3 + 2 \\ 0 & = y^{2} - 5 y - y + 9 - 3 + 2 \\ 0 & = y^{2} - 6 y + 9 - 3 + 2 \\ 0 & = y^{2} - 6 y + 6 + 2 \\ 0 & = y^{2} - 6 y + 8 \end{aligned}

$\begin{aligned} y &= \left(y - 3\right)^2+\left(y - 3\right)+2 \\ 0 &= {{{{{\left({{y}-{3}}\right)}^{2}}+{\left({{y}-{3}}\right)}}+{2}}-{y}}\\0 &= {{{{{{{y}^{2}}-{{6}\,{}{y}}}+{9}}+{\left({{y}-{3}}\right)}}+{2}}-{y}}\\0 &= {{{{{{{{y}^{2}}-{{6}\,{}{y}}}+{9}}+{y}}-{3}}+{2}}-{y}}\\0 &= {{{{{{{{y}^{2}}-{{6}\,{}{y}}}+{y}}-{y}}+{9}}-{3}}+{2}}\\0 &= {{{{{{{y}^{2}}-{{5}\,{}{y}}}-{y}}+{9}}-{3}}+{2}}\\0 &= {{{{{{y}^{2}}-{{6}\,{}{y}}}+{9}}-{3}}+{2}}\\0 &= {{{{{y}^{2}}-{{6}\,{}{y}}}+{6}}+{2}}\\0 &= {{{{y}^{2}}-{{6}\,{}{y}}}+{8}} \end{aligned}$

STEP: Solve the equation

0 = y^{2} - 6 y + 8

$0 = {{{{y}^{2}}-{{6}\,{}{y}}}+{8}}$
[−2 points ⇒ 2 / 6 points left]

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{6}$ . This is the coefficient of the middle term. The product of the two numbers must be $\text{8}$ . This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{8}$ . The two numbers are $-\text{2}$ and $-\text{4}$ . We will re-write the middle term, forming two terms, using these two numbers.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = y^{2} - 6 y + 8 \\ 0 & = y^{2} - 2 y - 4 y + 8 \end{aligned}

$\begin{aligned} 0 &= {{{{y}^{2}}-{{6}\,{}{y}}}+{8}} \\ 0 &= y^{2}-2 y-4 y+8 \end{aligned}$

We now group the first two terms together and group the last terms together and factorise.

\begin{aligned} 0 & = y^{2} - 2 y - 4 y + 8 \\ 0 & = - y (- y + 2) + 4 (- y + 2) \end{aligned}

$\begin{aligned} 0 &= \color{red}{y^{2}-2 y}\color{blue}{-4 y+8}\\ 0 &= \color{red}{- y(- y+2)}\color{blue}{+4(- y+2)} \end{aligned}$

The last thing now is to factor the highest common factor.

\begin{aligned} 0 & = (- y + 4) (- y + 2) \end{aligned}

$\begin{aligned} 0 &= (- y+4)(- y+2) \end{aligned}$

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $y$ -values of our solutions. The $y$ -values are 4 and 2.

STEP: Calculate the corresponding values of

x

$x$
[−2 points ⇒ 0 / 6 points left]

We will substitute each $y$ -value into the linear equation. This will give us the corresponding values of $x$ .

NOTE: We can do that by substituting

y

$y$ into any of the original equations. It is easier to use the linear one.

For $y = 4$ :

\begin{aligned} x & = y - 3 \\ = 1 \end{aligned}

$\begin{aligned} x &= y - 3 \\ &= 1 \end{aligned}$

For $y = 2$ :

\begin{aligned} x & = y - 3 \\ = - 1 \end{aligned}

$\begin{aligned} x &= y - 3 \\ &= -1 \end{aligned}$

The solutions are $x = 1$ together with $y = 4$ , and $x = -1$ with $y = 2$ . In coordinates pairs, the solutions are $\left(1; 4\right)$ and $\left(-1; 2\right)$ .

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(1; 4\right)$ and $\left(-1; 2\right)$ . These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(1; 4\right)$ and $\left(-1; 2\right)$ .

Submit your answer as: and

Simultaneous equations: intersection points

The graph below shows the equations:

\begin{matrix} y = - \frac{8 x}{3} + \frac{32}{3} and y = - \frac{4 x^{2}}{3} + 4 x + \frac{16}{3} \end{matrix}

$\begin{array}{c}y = - \frac{8 x}{3} + \frac{32}{3} \quad \text{and} \quad y = - \frac{4 x^{2}}{3} + 4 x + \frac{16}{3}\end{array}$

Determine the points of intersection for these two curves.

INSTRUCTION:

Write your answers as coordinate pairs like this: ( 3; -4) and ( 1; 8).
It does not matter which coordinate pair you type first.
Do not round your answers.

Answer: The points of intersection are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Solve the equations simultaneously to find their intersection points.

STEP: Combine the equations by substitution
[−1 point ⇒ 5 / 6 points left]

We can substitute immediately: we know that for the linear equation $y$ is equal to the quantity $- \frac{8 x}{3} + \frac{32}{3}$ ; and at the same time, the quadratic equation also contains the value $y$ . Substitute like this:

\begin{aligned} y & is - \frac{8 x}{3} + \frac{32}{3} \\ and: \\ y & is also - \frac{4 x^{2}}{3} + 4 x + \frac{16}{3} \\ Therefore: \\ - \frac{8 x}{3} + \frac{32}{3} & = - \frac{4 x^{2}}{3} + 4 x + \frac{16}{3} \end{aligned}

$\begin{aligned} y & \; \text{ is } \; - \frac{8 x}{3} + \frac{32}{3} \\ \text{and:} \quad & \\ y & \; \text {is also } \; - \frac{4 x^{2}}{3} + 4 x + \frac{16}{3} \\ \text{Therefore: } \quad & \\ - \frac{8 x}{3} + \frac{32}{3} & = - \frac{4 x^{2}}{3} + 4 x + \frac{16}{3} \end{aligned}$

STEP: Arrange the equation in standard form
[−2 points ⇒ 3 / 6 points left]

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is $\text{3}$ . So multiply both sides of the equation by $\text{3}$ in order to cancel the denominators. Then arrange the equation in standard form.

\begin{aligned} (3) (- \frac{8 x}{3} + \frac{32}{3}) & = (- \frac{4 x^{2}}{3} + 4 x + \frac{16}{3}) (3) \\ - 8 x + 32 & = - 4 x^{2} + 12 x + 16 \\ 0 & = - 4 x^{2} + 20 x - 16 \end{aligned}

$\begin{aligned} (\text{3})\left(- \frac{8 x}{3} + \frac{32}{3}\right) & = \left(- \frac{4 x^{2}}{3} + 4 x + \frac{16}{3}\right)(\text{3}) \\ - 8 x + 32 & = - 4 x^{2} + 12 x + 16 \\ 0 & = - 4 x^{2} + 20 x - 16 \end{aligned}$

In this case, there is a common factor of $\text{−4}$ for all of the terms. Divide out this factor before factorising.

\begin{matrix} 0 = x^{2} - 5 x + 4 \end{matrix}

$\begin{array}{c}0 = x^{2} - 5 x + 4\end{array}$

STEP: Solve the equation
[−1 point ⇒ 2 / 6 points left]

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two $x$ -values.

\begin{aligned} 0 & = x^{2} - 5 x + 4 \\ 0 & = (x - 4) (x - 1) \\ x = 1 & and x = 4 \end{aligned}

$\begin{aligned} 0 & = x^{2} - 5 x + 4 \\ 0 & = \left(x - 4\right) \left(x - 1\right) \\ x = 1 & \quad \textrm{ and } \quad x = 4 \end{aligned}$

STEP: Use the values of

x

$x$ to find the values of

y

$y$
[−2 points ⇒ 0 / 6 points left]

Great... but it isn't over yet. We must now find the values of $y$ for these equations. Substitute in the values $x = 1$ and $x = 4$ into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

\begin{aligned} If x = 1 : \\ y & = - \frac{8 (1)}{3} + \frac{32}{3} \\ y & = 8 \\ If x = 4 : \\ y & = - \frac{8 (4)}{3} + \frac{32}{3} \\ y & = 0 \end{aligned}

$\begin{aligned} \text{If } \; x = 1: \; & \\ y & = - \frac{8 (1)}{3} + \frac{32}{3} \\ y & = 8 \\ \text{If } \; x = 4: \; & \\ y & = - \frac{8 (4)}{3} + \frac{32}{3} \\ y & = 0 \end{aligned}$

Finally! The solutions are $\left(1 ; 8\right)$ and $\left(4 ; 0\right)$ .

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are $\left(1 ; 8\right)$ and $\left(4 ; 0\right)$ .

Submit your answer as: and

Simultaneous equations: intersection points

The graph below shows the equations:

\begin{matrix} y = - \frac{8 x}{3} and y = \frac{4 x^{2}}{3} - \frac{4 x}{3} - 16 \end{matrix}

$\begin{array}{c}y = - \frac{8 x}{3} \quad \text{and} \quad y = \frac{4 x^{2}}{3} - \frac{4 x}{3} - 16\end{array}$

Determine the points of intersection for these two curves.

INSTRUCTION:

Write your answers as coordinate pairs like this: ( 3; -4) and ( 1; 8).
It does not matter which coordinate pair you type first.
Do not round your answers.

Answer: The points of intersection are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Solve the equations simultaneously to find their intersection points.

STEP: Combine the equations by substitution
[−1 point ⇒ 5 / 6 points left]

We can substitute immediately: we know that for the linear equation $y$ is equal to the quantity $- \frac{8 x}{3}$ ; and at the same time, the quadratic equation also contains the value $y$ . Substitute like this:

\begin{aligned} y & is - \frac{8 x}{3} \\ and: \\ y & is also \frac{4 x^{2}}{3} - \frac{4 x}{3} - 16 \\ Therefore: \\ - \frac{8 x}{3} & = \frac{4 x^{2}}{3} - \frac{4 x}{3} - 16 \end{aligned}

$\begin{aligned} y & \; \text{ is } \; - \frac{8 x}{3} \\ \text{and:} \quad & \\ y & \; \text {is also } \; \frac{4 x^{2}}{3} - \frac{4 x}{3} - 16 \\ \text{Therefore: } \quad & \\ - \frac{8 x}{3} & = \frac{4 x^{2}}{3} - \frac{4 x}{3} - 16 \end{aligned}$

STEP: Arrange the equation in standard form
[−2 points ⇒ 3 / 6 points left]

\begin{aligned} (3) (- \frac{8 x}{3}) & = (\frac{4 x^{2}}{3} - \frac{4 x}{3} - 16) (3) \\ - 8 x & = 4 x^{2} - 4 x - 48 \\ 0 & = 4 x^{2} + 4 x - 48 \end{aligned}

$\begin{aligned} (\text{3})\left(- \frac{8 x}{3}\right) & = \left(\frac{4 x^{2}}{3} - \frac{4 x}{3} - 16\right)(\text{3}) \\ - 8 x & = 4 x^{2} - 4 x - 48 \\ 0 & = 4 x^{2} + 4 x - 48 \end{aligned}$

In this case, there is a common factor of $\text{4}$ for all of the terms. Divide out this factor before factorising.

\begin{matrix} 0 = x^{2} + x - 12 \end{matrix}

$\begin{array}{c}0 = x^{2} + x - 12\end{array}$

STEP: Solve the equation
[−1 point ⇒ 2 / 6 points left]

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two $x$ -values.

\begin{aligned} 0 & = x^{2} + x - 12 \\ 0 & = (x - 3) (x + 4) \\ x = - 4 & and x = 3 \end{aligned}

$\begin{aligned} 0 & = x^{2} + x - 12 \\ 0 & = \left(x - 3\right) \left(x + 4\right) \\ x = -4 & \quad \textrm{ and } \quad x = 3 \end{aligned}$

STEP: Use the values of

x

$x$ to find the values of

y

$y$
[−2 points ⇒ 0 / 6 points left]

Great... but it isn't over yet. We must now find the values of $y$ for these equations. Substitute in the values $x = -4$ and $x = 3$ into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

\begin{aligned} If x = - 4 : \\ y & = - \frac{8 (- 4)}{3} \\ y & = \frac{32}{3} \\ If x = 3 : \\ y & = - \frac{8 (3)}{3} \\ y & = - 8 \end{aligned}

$\begin{aligned} \text{If } \; x = -4: \; & \\ y & = - \frac{8 (-4)}{3} \\ y & = \frac{32}{3} \\ \text{If } \; x = 3: \; & \\ y & = - \frac{8 (3)}{3} \\ y & = -8 \end{aligned}$

Finally! The solutions are $\left(-4 ; \frac{32}{3}\right)$ and $\left(3 ; -8\right)$ .

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are $\left(-4 ; \frac{32}{3}\right)$ and $\left(3 ; -8\right)$ .

Submit your answer as: and

Simultaneous equations: intersection points

The graph below shows the equations:

\begin{matrix} y = \frac{4 x}{3} - \frac{4}{3} and y = - \frac{4 x^{2}}{3} - \frac{20 x}{3} - 8 \end{matrix}

$\begin{array}{c}y = \frac{4 x}{3} - \frac{4}{3} \quad \text{and} \quad y = - \frac{4 x^{2}}{3} - \frac{20 x}{3} - 8\end{array}$

Find the intersection points of the curves.

INSTRUCTION:

Write your answers as coordinate pairs like this: ( 3; -4) and ( 1; 8).
It does not matter which coordinate pair you type first.
Do not round your answers.

Answer: The points of intersection are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Solve the equations simultaneously to find their intersection points.

STEP: Combine the equations by substitution
[−1 point ⇒ 5 / 6 points left]

We can substitute immediately: we know that for the linear equation $y$ is equal to the quantity $\frac{4 x}{3} - \frac{4}{3}$ ; and at the same time, the quadratic equation also contains the value $y$ . Substitute like this:

\begin{aligned} y & is \frac{4 x}{3} - \frac{4}{3} \\ and: \\ y & is also - \frac{4 x^{2}}{3} - \frac{20 x}{3} - 8 \\ Therefore: \\ \frac{4 x}{3} - \frac{4}{3} & = - \frac{4 x^{2}}{3} - \frac{20 x}{3} - 8 \end{aligned}

$\begin{aligned} y & \; \text{ is } \; \frac{4 x}{3} - \frac{4}{3} \\ \text{and:} \quad & \\ y & \; \text {is also } \; - \frac{4 x^{2}}{3} - \frac{20 x}{3} - 8 \\ \text{Therefore: } \quad & \\ \frac{4 x}{3} - \frac{4}{3} & = - \frac{4 x^{2}}{3} - \frac{20 x}{3} - 8 \end{aligned}$

STEP: Arrange the equation in standard form
[−2 points ⇒ 3 / 6 points left]

\begin{aligned} (3) (\frac{4 x}{3} - \frac{4}{3}) & = (- \frac{4 x^{2}}{3} - \frac{20 x}{3} - 8) (3) \\ 4 x - 4 & = - 4 x^{2} - 20 x - 24 \\ 0 & = - 4 x^{2} - 24 x - 20 \end{aligned}

$\begin{aligned} (\text{3})\left(\frac{4 x}{3} - \frac{4}{3}\right) & = \left(- \frac{4 x^{2}}{3} - \frac{20 x}{3} - 8\right)(\text{3}) \\ 4 x - 4 & = - 4 x^{2} - 20 x - 24 \\ 0 & = - 4 x^{2} - 24 x - 20 \end{aligned}$

In this case, there is a common factor of $\text{−4}$ for all of the terms. Divide out this factor before factorising.

\begin{matrix} 0 = x^{2} + 6 x + 5 \end{matrix}

$\begin{array}{c}0 = x^{2} + 6 x + 5\end{array}$

STEP: Solve the equation
[−1 point ⇒ 2 / 6 points left]

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two $x$ -values.

\begin{aligned} 0 & = x^{2} + 6 x + 5 \\ 0 & = (x + 1) (x + 5) \\ x = - 5 & and x = - 1 \end{aligned}

$\begin{aligned} 0 & = x^{2} + 6 x + 5 \\ 0 & = \left(x + 1\right) \left(x + 5\right) \\ x = -5 & \quad \textrm{ and } \quad x = -1 \end{aligned}$

STEP: Use the values of

x

$x$ to find the values of

y

$y$
[−2 points ⇒ 0 / 6 points left]

Great... but it isn't over yet. We must now find the values of $y$ for these equations. Substitute in the values $x = -5$ and $x = -1$ into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

\begin{aligned} If x = - 5 : \\ y & = \frac{4 (- 5)}{3} - \frac{4}{3} \\ y & = - 8 \\ If x = - 1 : \\ y & = \frac{4 (- 1)}{3} - \frac{4}{3} \\ y & = - \frac{8}{3} \end{aligned}

$\begin{aligned} \text{If } \; x = -5: \; & \\ y & = \frac{4 (-5)}{3} - \frac{4}{3} \\ y & = -8 \\ \text{If } \; x = -1: \; & \\ y & = \frac{4 (-1)}{3} - \frac{4}{3} \\ y & = - \frac{8}{3} \end{aligned}$

Finally! The solutions are $\left(-5 ; -8\right)$ and $\left(-1 ; - \frac{8}{3}\right)$ .

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are $\left(-5 ; -8\right)$ and $\left(-1 ; - \frac{8}{3}\right)$ .

Submit your answer as: and

Substitution with quadratic equations

Here are two equations, which we can solve simultaneously using substitution:

\begin{aligned} y & = x^{2} + x - 7 \\ y & = 2 x - 1 \end{aligned}

$\begin{aligned} y &= x^{2} + x - 7 \\ y & = 2 x - 1 \end{aligned}$

Which of the following equations shows the correct substitution step?

A	$2 x - 1=y^{2} + y - 7$
B	$y=(2 x)^2+(2 x)-7$
C	$2 y - 1=x^{2} + x - 7$
D	$2 x - 1=x^{2} + x - 7$

Answer: The correct substitution step is choice: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to substitute $2 x - 1$ in place of $y$ in the quadratic equation $y = x^{2} + x - 7$ .

STEP: Substitute the linear equation into the quadratic equation
[−1 point ⇒ 0 / 1 points left]

We have two equations that we will solve by substitution. The equations are:

\begin{aligned} quadratic equation ⟶ & y = x^{2} + x - 7 \\ linear equation ⟶ & y = 2 x - 1 \end{aligned}

$\begin{aligned} {\text{quadratic equation}} \quad \longrightarrow \quad & y = x^{2} + x - 7 \\ {\text{linear equation}} \quad \longrightarrow \quad & y = 2 x - 1 \end{aligned}$

We need to substitute $2 x - 1$ from the linear equation in place of $y$ in the quadratic equation.

\begin{matrix} 2 x - 1 = x^{2} + x - 7 \end{matrix}

$\begin{array}{c}\color{red}{2 x - 1}=x^{2} + x - 7\end{array}$

NOTE: We could also substitute the value of

y

$y$ from the quadratic equation into the linear equation. The solution for this question did not consider this case.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for $x$ . The question does not ask us to solve the simultaneous equations.

The correct substitution step is $2 x - 1=x^{2} + x - 7$ which is choice D.

Submit your answer as:

Substitution with quadratic equations

Here are two equations, which we can solve simultaneously using substitution:

\begin{aligned} y & = x^{2} - x + 5 \\ y & = - 2 x + 7 \end{aligned}

$\begin{aligned} y &= x^{2} - x + 5 \\ y & = - 2 x + 7 \end{aligned}$

Which of the following equations shows the correct substitution step?

A	$- 2 x + 7=x^{2} - x + 5$
B	$- 2 x + 7=y^{2} - y + 5$
C	$y=(- 2 x + 7)^2-(- 2 x + 7)+5$
D	$- 2 x + 7=x^{2} - x$

Answer: The correct substitution step is choice: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to substitute $- 2 x + 7$ in place of $y$ in the quadratic equation $y = x^{2} - x + 5$ .

STEP: Substitute the linear equation into the quadratic equation
[−1 point ⇒ 0 / 1 points left]

We have two equations that we will solve by substitution. The equations are:

\begin{aligned} quadratic equation ⟶ & y = x^{2} - x + 5 \\ linear equation ⟶ & y = - 2 x + 7 \end{aligned}

$\begin{aligned} {\text{quadratic equation}} \quad \longrightarrow \quad & y = x^{2} - x + 5 \\ {\text{linear equation}} \quad \longrightarrow \quad & y = - 2 x + 7 \end{aligned}$

We need to substitute $- 2 x + 7$ from the linear equation in place of $y$ in the quadratic equation.

\begin{matrix} - 2 x + 7 = x^{2} - x + 5 \end{matrix}

$\begin{array}{c}\color{red}{- 2 x + 7}=x^{2} - x + 5\end{array}$

NOTE: We could also substitute the value of

y

$y$ from the quadratic equation into the linear equation. The solution for this question did not consider this case.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for $x$ . The question does not ask us to solve the simultaneous equations.

The correct substitution step is $- 2 x + 7=x^{2} - x + 5$ which is choice A.

Submit your answer as:

Substitution with quadratic equations

Here are two equations, which we can solve simultaneously using substitution:

\begin{aligned} y & = x^{2} - 2 x - 4 \\ y & = 3 x - 8 \end{aligned}

$\begin{aligned} y &= x^{2} - 2 x - 4 \\ y & = 3 x - 8 \end{aligned}$

Which of the following equations shows the correct substitution step?

A	$y=(3 x)^2-2(3 x)-4$
B	$3 x - 8=x^{2} - 2 x - 4$
C	$y=(3 x - 8)^2-2(3 x - 8)-4$
D	$3 x - 8=y^{2} - 2 y - 4$

Answer: The correct substitution step is choice: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

You need to substitute $3 x - 8$ in place of $y$ in the quadratic equation $y = x^{2} - 2 x - 4$ .

STEP: Substitute the linear equation into the quadratic equation
[−1 point ⇒ 0 / 1 points left]

We have two equations that we will solve by substitution. The equations are:

\begin{aligned} quadratic equation ⟶ & y = x^{2} - 2 x - 4 \\ linear equation ⟶ & y = 3 x - 8 \end{aligned}

$\begin{aligned} {\text{quadratic equation}} \quad \longrightarrow \quad & y = x^{2} - 2 x - 4 \\ {\text{linear equation}} \quad \longrightarrow \quad & y = 3 x - 8 \end{aligned}$

We need to substitute $3 x - 8$ from the linear equation in place of $y$ in the quadratic equation.

\begin{matrix} 3 x - 8 = x^{2} - 2 x - 4 \end{matrix}

$\begin{array}{c}\color{red}{3 x - 8}=x^{2} - 2 x - 4\end{array}$

NOTE: We could also substitute the value of

y

$y$ from the quadratic equation into the linear equation. The solution for this question did not consider this case.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for $x$ . The question does not ask us to solve the simultaneous equations.

The correct substitution step is $3 x - 8=x^{2} - 2 x - 4$ which is choice B.

Submit your answer as:

Equations with fractions

Solve the following equations simultaneously to find the values of $x$ and $y$ :

\begin{aligned} \frac{4}{x} - \frac{7}{x y} & = \frac{4}{y} x, y \neq 0 \\ - 4 y & = - 2 x - 5 \end{aligned}

$\begin{aligned} \frac{4}{x} - \frac{7}{x y} &= \frac{4}{y} \qquad x, y \neq 0 \\ - 4 y &= - 2 x - 5 \end{aligned}$

Your answer should be exact (do not round off).

Answer:

The solution is $x =$ and $y =$ .

numeric

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The first thing you should do is rearrange the first equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is $xy$ .

STEP: Rearrange the first equation
[−1 point ⇒ 4 / 5 points left]

We need to solve the equations simultaneously. To do that, we should start by rearranging the first equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by $xy$ , which is the LCD of the fractions. That will cancel all of the denominators.

\begin{aligned} \frac{4}{x} - \frac{7}{x y} & = \frac{4}{y} \\ x y (\frac{4}{x} - \frac{7}{x y}) & = x y (\frac{4}{y}) \\ \frac{4}{x} x y - 7 & = \frac{4}{y} x y \\ 4 y - 7 & = 4 x \end{aligned}

$\begin{aligned} \frac{4}{x} - \frac{7}{x y} &= \frac{4}{y} \\ x y\left(\frac{4}{x} - \frac{7}{x y}\right)&=x y\left(\frac{4}{y}\right) \\ \frac{4}{x} x y - 7 &= \frac{4}{y} x y \\ 4 y - 7 &= 4 x \end{aligned}$

STEP: Pick a method and solve for whichever variable comes more easily
[−3 points ⇒ 1 / 5 points left]

Now we are ready to solve the following equations:

\begin{aligned} 4 y - 7 = 4 x \\ - 4 y = - 2 x - 5 \end{aligned}

$\begin{aligned} 4 y - 7 = 4 x \\ - 4 y = - 2 x - 5 \end{aligned}$

These equations invite elimination, because the $y$ -terms have equal and opposite coefficients. We can eliminate the $y$ -terms by adding the equations.

Eliminate the $y$ and solve for $x$ :

\begin{aligned} 4 y - 7 & = 4 x \\ _{second equation}^{Add the} \to + (- 4 y) & = + (- 2 x - 5) \\ - 7 - 4 y + 4 y & = - 5 - 2 x + 4 x \\ - 7 & = 2 x - 5 \\ - 2 x & = 2 \\ x & = - 1 \end{aligned}

$\begin{aligned} 4 y - 7 & = 4 x \\ ^{\text{Add the}}_{\text{second equation}} \quad \rightarrow \quad +\left(- 4 y\right)& =+\left(- 2 x - 5\right) \\ \hline - 7 - 4 y + 4 y & = - 5 - 2 x + 4 x \\ -7 & = 2 x - 5 \\ - 2 x & = 2 \\ x & = -1 \end{aligned}$

Great - we have the first value, $x = -1$ .

STEP: Solve for the other variable
[−1 point ⇒ 0 / 5 points left]

Now the last step: find the value of $y$ . We can do this using either of the equations in the question: but the second eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the first equation).

\begin{aligned} - 4 y & = - 2 x - 5 \\ - 4 y & = - 2 (- 1) - 5 \\ - 4 y & = - 3 \\ y & = \frac{3}{4} \end{aligned}

$\begin{aligned} - 4 y & = - 2 x - 5 \\ - 4 y & = - 2 \left(-1\right) - 5 \\ - 4 y & = -3 \\ y & = \frac{3}{4} \end{aligned}$

The answer is the pair of numbers $x = -1$ and $y = \frac{3}{4}$ . As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

NOTE: Remember, the fractions force

x, y \neq 0

$x, y \neq 0$ , which is why there are open intervals at the

x

$x$ - and

y

$y$ -intercepts of the fraction equation.

The values which solve these two equations are $x = -1$ and $y = \frac{3}{4}$ .

Submit your answer as: and

Equations with fractions

Solve the following equations simultaneously to find the values of $x$ and $y$ :

\begin{aligned} \frac{1}{y} & = \frac{4}{x} - \frac{9}{x y} x, y \neq 0 \\ - 2 x - 15 & = - 4 y \end{aligned}

$\begin{aligned} \frac{1}{y} &= \frac{4}{x} - \frac{9}{x y} \qquad x, y \neq 0 \\ - 2 x - 15 &= - 4 y \end{aligned}$

Your answer should be exact (do not round off).

Answer:

The solution is $x =$ and $y =$ .

numeric

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The first thing you should do is rearrange the first equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is $xy$ .

STEP: Rearrange the first equation
[−1 point ⇒ 4 / 5 points left]

\begin{aligned} \frac{1}{y} & = \frac{4}{x} - \frac{9}{x y} \\ x y (\frac{1}{y}) & = x y (\frac{4}{x} - \frac{9}{x y}) \\ \frac{x y}{y} & = \frac{4}{x} x y - 9 \\ x & = 4 y - 9 \end{aligned}

$\begin{aligned} \frac{1}{y} &= \frac{4}{x} - \frac{9}{x y} \\ x y\left(\frac{1}{y}\right)&=x y\left(\frac{4}{x} - \frac{9}{x y}\right) \\ \frac{x y}{y} &= \frac{4}{x} x y - 9 \\ x &= 4 y - 9 \end{aligned}$

STEP: Pick a method and solve for whichever variable comes more easily
[−3 points ⇒ 1 / 5 points left]

Now we are ready to solve the following equations:

\begin{aligned} x = 4 y - 9 \\ - 2 x - 15 = - 4 y \end{aligned}

$\begin{aligned} x = 4 y - 9 \\ - 2 x - 15 = - 4 y \end{aligned}$

These equations invite elimination, because the $y$ -terms have equal and opposite coefficients. We can eliminate the $y$ -terms by adding the equations.

Eliminate the $y$ and solve for $x$ :

\begin{aligned} x & = 4 y - 9 \\ _{second equation}^{Add the} \to + (- 2 x - 15) & = + (- 4 y) \\ - 15 - 2 x + x & = - 9 - 4 y + 4 y \\ - x - 15 & = - 9 \\ - x & = 6 \\ x & = - 6 \end{aligned}

$\begin{aligned} x & = 4 y - 9 \\ ^{\text{Add the}}_{\text{second equation}} \quad \rightarrow \quad +\left(- 2 x - 15\right)& =+\left(- 4 y\right) \\ \hline - 15 - 2 x + x & = - 9 - 4 y + 4 y \\ - x - 15 & = -9 \\ - x & = 6 \\ x & = -6 \end{aligned}$

Great - we have the first value, $x = -6$ .

STEP: Solve for the other variable
[−1 point ⇒ 0 / 5 points left]

\begin{aligned} - 2 x - 15 & = - 4 y \\ - 2 (- 6) - 15 & = - 4 y \\ 4 y & = 3 \\ y & = \frac{3}{4} \end{aligned}

$\begin{aligned} - 2 x - 15 & = - 4 y \\ - 2 \left(-6\right) - 15 & = - 4 y \\ 4 y & = 3 \\ y & = \frac{3}{4} \end{aligned}$

The answer is the pair of numbers $x = -6$ and $y = \frac{3}{4}$ . As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

NOTE: Remember, the fractions force

x, y \neq 0

$x, y \neq 0$ , which is why there are open intervals at the

x

$x$ - and

y

$y$ -intercepts of the fraction equation.

The values which solve these two equations are $x = -6$ and $y = \frac{3}{4}$ .

Submit your answer as: and

Equations with fractions

What are the values of $x$ and $y$ which solve these equations simultaneously?

\begin{aligned} \frac{6}{x y} & = \frac{6}{y} - \frac{1}{x} x, y \neq 0 \\ 4 x & = 6 y - 4 \end{aligned}

$\begin{aligned} \frac{6}{x y} &= \frac{6}{y} - \frac{1}{x} \qquad x, y \neq 0 \\ 4 x &= 6 y - 4 \end{aligned}$

Your answer should be exact (do not round off).

Answer:

The solution is $x =$ and $y =$ .

numeric

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

The first thing you should do is rearrange the first equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is $xy$ .

STEP: Rearrange the first equation
[−1 point ⇒ 4 / 5 points left]

\begin{aligned} \frac{6}{x y} & = \frac{6}{y} - \frac{1}{x} \\ x y (\frac{6}{x y}) & = x y (\frac{6}{y} - \frac{1}{x}) \\ 6 & = - \frac{1}{x} x y + \frac{6}{y} x y \\ 6 & = 6 x - y \end{aligned}

$\begin{aligned} \frac{6}{x y} &= \frac{6}{y} - \frac{1}{x} \\ x y\left(\frac{6}{x y}\right)&=x y\left(\frac{6}{y} - \frac{1}{x}\right) \\ 6 &= - \frac{1}{x} x y + \frac{6}{y} x y \\ 6 &= 6 x - y \end{aligned}$

STEP: Pick a method and solve for whichever variable comes more easily
[−3 points ⇒ 1 / 5 points left]

Now we are ready to solve the following equations:

\begin{aligned} 6 = 6 x - y \\ 4 x = 6 y - 4 \end{aligned}

$\begin{aligned} 6 = 6 x - y \\ 4 x = 6 y - 4 \end{aligned}$

These equations are not arranged nicely for elimination or substitution. We must change the equations somehow. In this case, we can make a small change that will allow us to use elimination: we can multiply the first equation by $\text{6}$ so that the $y$ -coefficients will be ready to cancel. Then we can eliminate them by adding the equations.

First modify things to set up the elimination:

\begin{aligned} 6 (6) & = 6 (6 x - y) \\ 36 & = 36 x - 6 y \end{aligned}

$\begin{aligned} 6 \left(6 \right) & = 6 \left(6 x - y \right) \\ 36 & = 36 x - 6 y \end{aligned}$

Now do the elimination and complete the solution to find $x$ .

\begin{aligned} 36 & = 36 x - 6 y \\ _{second equation}^{Add the} \to + (4 x) & = + (6 y - 4) \\ 36 + 4 x & = - 4 + 36 x - 6 y + 6 y \\ 4 x + 36 & = 36 x - 4 \\ - 32 x & = - 40 \\ x & = \frac{5}{4} \end{aligned}

$\begin{aligned} 36 & = 36 x - 6 y \\ ^{\text{Add the}}_{\text{second equation}} \quad \rightarrow \quad +\left(4 x\right)& =+\left(6 y - 4\right) \\ \hline 36 + 4 x & = - 4 + 36 x - 6 y + 6 y \\ 4 x + 36 & = 36 x - 4 \\ - 32 x & = -40 \\ x & = \frac{5}{4} \end{aligned}$

Great - we have the first value, $x = \frac{5}{4}$ .

STEP: Solve for the other variable
[−1 point ⇒ 0 / 5 points left]

\begin{aligned} 4 x & = 6 y - 4 \\ 4 (\frac{5}{4}) & = 6 y - 4 \\ - 6 y & = - 9 \\ y & = \frac{3}{2} \end{aligned}

$\begin{aligned} 4 x & = 6 y - 4 \\ 4 \left(\frac{5}{4}\right) & = 6 y - 4 \\ - 6 y & = -9 \\ y & = \frac{3}{2} \end{aligned}$

The answer is the pair of numbers $x = \frac{5}{4}$ and $y = \frac{3}{2}$ . As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

NOTE: Remember, the fractions force

x, y \neq 0

$x, y \neq 0$ , which is why there are open intervals at the

x

$x$ - and

y

$y$ -intercepts of the fraction equation.

The values which solve these two equations are $x = \frac{5}{4}$ and $y = \frac{3}{2}$ .

Submit your answer as: and

Simultaneous equations: special outcomes

Solve for $x$ and $y$ :

\begin{aligned} y & = 3 x^{2} - 5 x + 1 \\ y & = x - 2 \end{aligned}

$\begin{aligned} y & = 3 x^{2} - 5 x + 1 \\ y & = x - 2 \end{aligned}$

INSTRUCTION: You should enter your answer in the form of a coordinate pair, (x; y). An example of an acceptable answer is (2; 5). If there is no solution, type no solution.

Answer: The solution is .

coordinate

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to start by substituting $x - 2$ in place of $y$ in the quadratic equation and simplify. You must expect a special kind of solution.

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 3 / 5 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is $y = 3 x^{2} - 5 x + 1$ and the linear equation is $y = x - 2$ . This means we expect two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

NOTE: We could also substitute the value of

y

$y$ from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.

\begin{aligned} x - 2 & = 3 x^{2} - 5 x + 1 \\ 0 & = 3 x^{2} - 5 x + 1 - x + 2 \\ 0 & = 3 x^{2} - 5 x - x + 1 + 2 \\ 0 & = 3 x^{2} - 6 x + 3 \end{aligned}

$\begin{aligned} x - 2 &= 3 x^{2} - 5 x + 1 \\ 0 &= 3 x^{2} - 5 x + 1-x+2\\ 0 &= 3 x^{2}-5 x-x+1+2\\ 0 &= 3 x^{2}-6 x+3 \end{aligned}$

STEP: Solve the equation

0 = 3 x^{2} - 6 x + 3

$0=3 x^{2}-6 x+3$
[−2 points ⇒ 1 / 5 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE:We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

We will factorise using the grouping method. For this method, we need two numbers that add to give $-\text{6}$ . This is the coefficient of the middle term. The product of the two numbers must be $\text{9}$ . This is the product of the coefficient of the first term $\text{3}$ and the value of the last term $\text{3}$ . If we take $-\text{3}$ which is half of the value of $-\text{6}$ (the coefficient of the middle term) and square it, we get the product $\text{9}$ . In this case we will use only one number to factorise, that is, $-\text{3}$ . We will re-write the middle term, forming two terms, using this value.

\begin{aligned} 0 & = 3 x^{2} - 6 x + 3 \\ 0 & = 3 x^{2} - 3 x - 3 x + 3 \end{aligned}

$\begin{aligned} 0 &= 3 x^{2}-6 x+3 \\ 0 &= 3 x^{2}-3 x-3 x+3 \end{aligned}$

We now group the first two terms together and group the last terms together and factorise.

\begin{aligned} 0 & = 3 x^{2} - 3 x - 3 x + 3 \\ 0 & = 3 x (x - 1) - 3 (x - 1) \end{aligned}

$\begin{aligned} 0 &= \color{red}{3 x^{2}-3 x}\color{blue}{-3 x+3} \\ 0 &= \color{red}{3 x(x - 1)}\color{blue}{ - 3(x - 1)} \end{aligned}$

The last thing now is to factor the highest common factor.

\begin{matrix} 0 = (3 x - 3) (x - 1) \end{matrix}

$\begin{array}{c}0 = (3 x - 3)(x - 1)\end{array}$

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. In this case we have one root, $\text{1}$ . This root is the $x$ -value of our solution. This $x$ -value will have one corresponding $y$ value.

STEP: Calculate the corresponding value of

y

$y$
[−1 point ⇒ 0 / 5 points left]

Now that we have the value of $x$ , we will substitute it into the linear equation to calculate the corresponding value of $y$ .

NOTE: We can do that by substituting

x

$x$ into any of the original equations. It is easier to use the linear one.

\begin{aligned} y & = x - 2 \\ = 1 (1) - 2 \\ = - 1 \end{aligned}

$\begin{aligned} y &= x - 2 \\ &= 1 \left( 1 \right) -2 \\ &= -1 \end{aligned}$

The solution is $x = 1$ together with $y = -1$ . Using coordinate pairs, the solution is $\left(1; -1\right)$ .

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(1; -1\right)$ . This is the same solution we have calculated. This means we are $\text{100}\%$ correct.

The solution is $\left(1; -1\right)$ .

Submit your answer as:

Simultaneous equations: special outcomes

Solve for $x$ and $y$ :

\begin{aligned} y & = 3 x^{2} + 5 x + 4 \\ y & = - 2 x - 3 \end{aligned}

$\begin{aligned} y & = 3 x^{2} + 5 x + 4 \\ y & = - 2 x - 3 \end{aligned}$

INSTRUCTION: You should enter your answer in the form of a coordinate pair, (x; y). An example of an acceptable answer is (2; 5). If there is no solution, type no solution.

Answer: The solution is .

one-of

type(string.nocase)

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to start by substituting $- 2 x - 3$ in place of $y$ in the quadratic equation and simplify. You must expect a special kind of solution.

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 2 / 4 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is $y = 3 x^{2} + 5 x + 4$ and the linear equation is $y = - 2 x - 3$ . This means we expect two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

NOTE: We could also substitute the value of

y

$y$ from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.

\begin{aligned} - 2 x - 3 & = 3 x^{2} + 5 x + 4 \\ 0 & = 3 x^{2} + 5 x + 4 + 2 x + 3 \\ 0 & = 3 x^{2} + 5 x + 2 x + 4 + 3 \\ 0 & = 3 x^{2} + 7 x + 7 \end{aligned}

$\begin{aligned} - 2 x - 3 &= 3 x^{2} + 5 x + 4 \\ 0 &= 3 x^{2} + 5 x + 4+2 x+3\\ 0 &= 3 x^{2}+5 x+2 x+4+3\\ 0 &= 3 x^{2}+7 x+7 \end{aligned}$

STEP: Solve the equation

0 = 3 x^{2} + 7 x + 7

$0=3 x^{2}+7 x+7$
[−2 points ⇒ 0 / 4 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

We need two numbers that add to give $\text{7}$ . This is the coefficient of the middle term. The product of the two numbers must be $\text{21}$ . This is the product of the coefficient of the first term $\text{3}$ and the value of the last term $\text{7}$ . In this case, there are no such numbers. We will now use the quadratic formula to solve this equation.

The solution for a quadratic equation of the form $ax^2 + bx + c = 0$ :

\begin{matrix} x = \frac{- b \pm \sqrt{(b^{2} - 4 a c)}}{2 a} \end{matrix}

$\begin{array}{c}x = \dfrac{-b \pm \sqrt{\left(b^2 - 4ac\right)}}{2a}\end{array}$

Before we can use the quadratic formula, we need to identify the coefficients $a$ , $b$ and $c$ from the equation $3 x^{2}+7 x+7 = 0$ These are:

\begin{matrix} a = 3, b = 7 and c = 7. \end{matrix}

$\begin{array}{c}a = 3, b = 7 \text{ and } c = 7.\end{array}$

We will substitute these values into the formula and simplify.

\begin{aligned} x & = \frac{- (7) \pm \sqrt{({(7)}^{2} - 4 (3) (7))}}{2 (3)} \\ = \frac{- 7 \pm \sqrt{(49 - (84))}}{6} \\ = \frac{- 7 \pm \sqrt{(- 35)}}{6} \end{aligned}

$\begin{aligned} x &= \dfrac{-\left(7\right) \pm \sqrt{\left(\left(7\right)^2 - 4\left(3\right) \left(7\right)\right)}} {2\left(3\right)} \\ &= \dfrac{-7 \pm \sqrt{\left(49 - \left(84\right)\right)}} {6} \\ &= \dfrac{-7 \pm \sqrt{\left(-35\right)}} {6} \end{aligned}$

The number under the square root sign is negative. This means that our quadratic equation has no real solution. This means that $y = 3 x^{2} + 5 x + 4$ and $y = - 2 x - 3$ cannot be solved simultaneously.

We can confirm our result by plotting the graph of each equation on the same set of axes. We do not expect these graphs to intersect.

Aha! The two graphs do not have an intersection point. This means that indeed the two equations have no solution.

The final answer is no solution.

Submit your answer as:

Simultaneous equations: special outcomes

Solve for $x$ and $y$ from the given equations:

\begin{aligned} y & = x^{2} - 2 x + 2 \\ y & = x - 5 \end{aligned}

$\begin{aligned} y & = x^{2} - 2 x + 2 \\ y & = x - 5 \end{aligned}$

INSTRUCTION: You should enter your answer in the form of a coordinate pair, (x; y). An example of an acceptable answer is (2; 5). If there is no solution, type no solution.

Answer: The solution is .

one-of

type(string.nocase)

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You need to start by substituting $x - 5$ in place of $y$ in the quadratic equation and simplify. You must expect a special kind of solution.

STEP: Substitute the linear equation into the quadratic equation
[−2 points ⇒ 2 / 4 points left]

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is $y = x^{2} - 2 x + 2$ and the linear equation is $y = x - 5$ . This means we expect two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

NOTE: We could also substitute the value of

y

$y$ from the quadratic equation into the linear equation. We could still get the same answer, but the calculations might become too long.

\begin{aligned} x - 5 & = x^{2} - 2 x + 2 \\ 0 & = x^{2} - 2 x + 2 - x + 5 \\ 0 & = x^{2} - 2 x - x + 2 + 5 \\ 0 & = x^{2} - 3 x + 7 \end{aligned}

$\begin{aligned} x - 5 &= x^{2} - 2 x + 2 \\ 0 &= x^{2} - 2 x + 2-x+5\\ 0 &= x^{2}-2 x-x+2+5\\ 0 &= x^{2}-3 x+7 \end{aligned}$

STEP: Solve the equation

0 = x^{2} - 3 x + 7

$0=x^{2}-3 x+7$
[−2 points ⇒ 0 / 4 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

We need two numbers that add to give $-\text{3}$ . This is the coefficient of the middle term. The product of the two numbers must be $\text{7}$ . This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{7}$ . In this case, there are no such numbers. We will now use the quadratic formula to solve this equation.

The solution for a quadratic equation of the form $ax^2 + bx + c = 0$ :

\begin{matrix} x = \frac{- b \pm \sqrt{(b^{2} - 4 a c)}}{2 a} \end{matrix}

$\begin{array}{c}x = \dfrac{-b \pm \sqrt{\left(b^2 - 4ac\right)}}{2a}\end{array}$

Before we can use the quadratic formula, we need to identify the coefficients $a$ , $b$ and $c$ from the equation $x^{2}-3 x+7 = 0$ These are:

\begin{matrix} a = 1, b = - 3 and c = 7. \end{matrix}

$\begin{array}{c}a = 1, b = -3 \text{ and } c = 7.\end{array}$

We will substitute these values into the formula and simplify.

\begin{aligned} x & = \frac{- (- 3) \pm \sqrt{({(- 3)}^{2} - 4 (1) (7))}}{2 (1)} \\ = \frac{3 \pm \sqrt{(9 - (28))}}{2} \\ = \frac{3 \pm \sqrt{(- 19)}}{2} \end{aligned}

$\begin{aligned} x &= \dfrac{-\left(-3\right) \pm \sqrt{\left(\left(-3\right)^2 - 4\left(1\right) \left(7\right)\right)}} {2\left(1\right)} \\ &= \dfrac{3 \pm \sqrt{\left(9 - \left(28\right)\right)}} {2} \\ &= \dfrac{3 \pm \sqrt{\left(-19\right)}} {2} \end{aligned}$

The number under the square root sign is negative. This means that our quadratic equation has no real solution. This means that $y = x^{2} - 2 x + 2$ and $y = x - 5$ cannot be solved simultaneously.

We can confirm our result by plotting the graph of each equation on the same set of axes. We do not expect these graphs to intersect.

Aha! The two graphs do not have an intersection point. This means that indeed the two equations have no solution.

The final answer is no solution.

Submit your answer as:

Simultaneous equations

Adapted from DBE Nov 2016 Grade 12, P1, Q1.3
Maths formulas

Solve for $x$ and $y$ :

\begin{matrix} x = - 4 y \\ and \\ x^{2} + 2 x y = 32 \end{matrix}

$\begin{array}{c} x = - 4 y \\ \text{and} \\ x^{2} + 2 x y= 32 \end{array}$

INSTRUCTIONS:

You should get two coordinate pairs in the form $(x_{1}; y_{1})$ and $(x_{2}; y_{2})$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The solutions are
and .

coordinate

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.

STEP: Substitute

x = - 4 y

$x = - 4 y$ into

x^{2} + 2 x y = 32

$x^{2} + 2 x y= 32$ and solve for

y

$y$
[−3 points ⇒ 1 / 4 points left]

We need to solve the following pair of equations for $x$ and $y$ :

\begin{matrix} x = - 4 y \\ and \\ x^{2} + 2 x y = 32 \end{matrix}

$\begin{array}{c} x = - 4 y \\ \text{and} \\ x^{2} + 2 x y= 32 \end{array}$

First, we can substitute $- 4 y$ in place of $x$ in the second equation:

\begin{aligned} (- 4 y)^{2} + 2 (- 4 y) (y) & = 32 \end{aligned}

$\begin{aligned} (-4y)^{2} + 2 (-4y) (y)& = 32 \end{aligned}$

We can simplify this equation to solve for $y$ :

\begin{aligned} (- 4 y)^{2} + 2 (- 4 y) (y) & = 32 \\ 16 y^{2} - 8 y^{2} & = 32 \\ 8 y^{2} & = 32 \\ y^{2} & = 4 \end{aligned}

$\begin{aligned} (-4y)^{2} + 2 (-4y) (y)& = 32 \\ 16 y^{2}-8 y^{2}& = 32 \\ 8 y^{2}& = 32 \\ y^{2}& = 4 \end{aligned}$

\begin{matrix} y = 2 or y = - 2 \end{matrix}

$\begin{array}{c} y = 2 \text{ or } y = -2 \end{array}$

STEP: Use the

y

$y$ -values to find the

x

$x$ -values
[−1 point ⇒ 0 / 4 points left]

Each of the $y$ -values that we have found will have a corresponding value of $x$ , which can be found from the equation $x = - 4 y$ .

For $y = 2$ :

\begin{aligned} x & = - 4 y \\ = - 4 (2) \\ = - 8 \end{aligned}

$\begin{aligned} x & = -4y \\ & = -4(2) \\ & = -8 \end{aligned}$

And for $y = -2$ :

\begin{aligned} x & = - 4 y \\ = - 4 (- 2) \\ = 8 \end{aligned}

$\begin{aligned} x & = -4y \\ & = -4(-2) \\ & = 8 \end{aligned}$

So the $x$ -values are $x = -8 \text{ or } x = 8$ .

Therefore the correct coordinate pairs are $(-8; 2)$ and $(8; -2)$ .

Submit your answer as: and

Simultaneous equations

Adapted from DBE Nov 2016 Grade 12, P1, Q1.3
Maths formulas

Solve for $x$ and $y$ :

\begin{matrix} x = 2 y \\ and \\ x^{2} - 4 x y = - 16 \end{matrix}

$\begin{array}{c} x = 2 y \\ \text{and} \\ x^{2} - 4 x y= -16 \end{array}$

INSTRUCTIONS:

You should get two coordinate pairs in the form $(x_{1}; y_{1})$ and $(x_{2}; y_{2})$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The solutions are
and .

coordinate

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.

STEP: Substitute

x = 2 y

$x = 2 y$ into

x^{2} - 4 x y = - 16

$x^{2} - 4 x y= -16$ and solve for

y

$y$
[−3 points ⇒ 1 / 4 points left]

We need to solve the following pair of equations for $x$ and $y$ :

\begin{matrix} x = 2 y \\ and \\ x^{2} - 4 x y = - 16 \end{matrix}

$\begin{array}{c} x = 2 y \\ \text{and} \\ x^{2} - 4 x y= -16 \end{array}$

First, we can substitute $2 y$ in place of $x$ in the second equation:

\begin{aligned} (2 y)^{2} - 4 (2 y) (y) & = - 16 \end{aligned}

$\begin{aligned} (2y)^{2} - 4 (2y) (y)& = -16 \end{aligned}$

We can simplify this equation to solve for $y$ :

\begin{aligned} (2 y)^{2} - 4 (2 y) (y) & = - 16 \\ 4 y^{2} - 8 y^{2} & = - 16 \\ - 4 y^{2} & = - 16 \\ y^{2} & = 4 \end{aligned}

$\begin{aligned} (2y)^{2} - 4 (2y) (y)& = -16 \\ 4 y^{2}-8 y^{2}& = -16 \\ - 4 y^{2}& = -16 \\ y^{2}& = 4 \end{aligned}$

\begin{matrix} y = 2 or y = - 2 \end{matrix}

$\begin{array}{c} y = 2 \text{ or } y = -2 \end{array}$

STEP: Use the

y

$y$ -values to find the

x

$x$ -values
[−1 point ⇒ 0 / 4 points left]

Each of the $y$ -values that we have found will have a corresponding value of $x$ , which can be found from the equation $x = 2 y$ .

For $y = 2$ :

\begin{aligned} x & = 2 y \\ = 2 (2) \\ = 4 \end{aligned}

$\begin{aligned} x & = 2y \\ & = 2(2) \\ & = 4 \end{aligned}$

And for $y = -2$ :

\begin{aligned} x & = 2 y \\ = 2 (- 2) \\ = - 4 \end{aligned}

$\begin{aligned} x & = 2y \\ & = 2(-2) \\ & = -4 \end{aligned}$

So the $x$ -values are $x = 4 \text{ or } x = -4$ .

Therefore the correct coordinate pairs are $(4; 2)$ and $(-4; -2)$ .

Submit your answer as: and

Simultaneous equations

Adapted from DBE Nov 2016 Grade 12, P1, Q1.3
Maths formulas

Solve for $x$ and $y$ :

\begin{matrix} x = - 2 y \\ and \\ x^{2} + 4 x y = - 36 \end{matrix}

$\begin{array}{c} x = - 2 y \\ \text{and} \\ x^{2} + 4 x y= -36 \end{array}$

INSTRUCTIONS:

You should get two coordinate pairs in the form $(x_{1}; y_{1})$ and $(x_{2}; y_{2})$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The solutions are
and .

coordinate

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.

STEP: Substitute

x = - 2 y

$x = - 2 y$ into

x^{2} + 4 x y = - 36

$x^{2} + 4 x y= -36$ and solve for

y

$y$
[−3 points ⇒ 1 / 4 points left]

We need to solve the following pair of equations for $x$ and $y$ :

\begin{matrix} x = - 2 y \\ and \\ x^{2} + 4 x y = - 36 \end{matrix}

$\begin{array}{c} x = - 2 y \\ \text{and} \\ x^{2} + 4 x y= -36 \end{array}$

First, we can substitute $- 2 y$ in place of $x$ in the second equation:

\begin{aligned} (- 2 y)^{2} + 4 (- 2 y) (y) & = - 36 \end{aligned}

$\begin{aligned} (-2y)^{2} + 4 (-2y) (y)& = -36 \end{aligned}$

We can simplify this equation to solve for $y$ :

\begin{aligned} (- 2 y)^{2} + 4 (- 2 y) (y) & = - 36 \\ 4 y^{2} - 8 y^{2} & = - 36 \\ - 4 y^{2} & = - 36 \\ y^{2} & = 9 \end{aligned}

$\begin{aligned} (-2y)^{2} + 4 (-2y) (y)& = -36 \\ 4 y^{2}-8 y^{2}& = -36 \\ - 4 y^{2}& = -36 \\ y^{2}& = 9 \end{aligned}$

\begin{matrix} y = 3 or y = - 3 \end{matrix}

$\begin{array}{c} y = 3 \text{ or } y = -3 \end{array}$

STEP: Use the

y

$y$ -values to find the

x

$x$ -values
[−1 point ⇒ 0 / 4 points left]

Each of the $y$ -values that we have found will have a corresponding value of $x$ , which can be found from the equation $x = - 2 y$ .

For $y = 3$ :

\begin{aligned} x & = - 2 y \\ = - 2 (3) \\ = - 6 \end{aligned}

$\begin{aligned} x & = -2y \\ & = -2(3) \\ & = -6 \end{aligned}$

And for $y = -3$ :

\begin{aligned} x & = - 2 y \\ = - 2 (- 3) \\ = 6 \end{aligned}

$\begin{aligned} x & = -2y \\ & = -2(-3) \\ & = 6 \end{aligned}$

So the $x$ -values are $x = -6 \text{ or } x = 6$ .

Therefore the correct coordinate pairs are $(-6; 3)$ and $(6; -3)$ .

Submit your answer as: and

Simultaneous equations with $xy$ terms

Determine the solution of the following equations:

\begin{aligned} - 2 y & = - 4 x - x y - 2 \\ y & = 2 x + 4 \end{aligned}

$\begin{aligned} - 2 y & = - 4 x-x y - 2\\ y & = 2 x + 4 \end{aligned}$

INSTRUCTION:

You should get two coordinate pairs in the form $\left(x_1; y_1\right)$ and $\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting $2 x + 4$ in place of $y$ in the hyperbolic equation and then solve for $x$ .

STEP: Substitute the linear equation into the hyperbolic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

\begin{aligned} hyperbolic equation ⟶ & - 2 y = - 4 x - x y - 2 \\ linear equation ⟶ & y = 2 x + 4 \end{aligned}

$\begin{aligned} {\text{hyperbolic equation}}\quad \longrightarrow \quad & - 2 y = - 4 x-x y - 2\\ {\text{linear equation}}\quad \longrightarrow \quad & y = 2 x + 4 \end{aligned}$

This means we want two pairs of $x$ and $y$ -values which solve both equations.

We will substitute the value of $y$ from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

NOTE: We could also isolate either

x

$x$ or

y

$y$ from

- 2 y = - 4 x - x y - 2

$- 2 y = - 4 x-x y - 2$ and substitute its value into

y = 2 x + 4

$y = 2 x + 4$ . We could still get the same answer, but the calculations might become too long.

\begin{aligned} - 2 (2 x + 4) & = - 4 x - x (2 x + 4) - 2 \\ 0 & = - 4 x - x (2 x + 4) - 2 + 2 (2 x + 4) \\ 0 & = - 4 x - (2 x^{2} + 4 x) - 2 + (4 x + 8) \\ 0 & = - 2 x^{2} - 8 x + 4 x - 2 + 8 \\ 0 & = - 2 x^{2} - 4 x - 2 + 8 \\ 0 & = - 2 x^{2} - 4 x + 6 \\ _{make the first term positive}^{divide each term by -1 to} \to 0 & = 2 x^{2} + 4 x - 6 \end{aligned}

$\begin{aligned} - 2 (2 x + 4) & = - 4 x-x (2 x + 4) - 2\\ 0 & = - 4 x-x (2 x + 4) - 2+2 (2 x + 4) \\ 0 & = {{{{-{{4}\,{}{x}}}-{\left({{{2}\,{}{{x}^{2}}}+{{4}\,{}{x}}}\right)}}-{2}}+{\left({{{4}\,{}{x}}+{8}}\right)}}\\0 & = {{{{{-{{2}\,{}{{x}^{2}}}}-{{8}\,{}{x}}}+{{4}\,{}{x}}}-{2}}+{8}}\\0 & = {{{{-{{2}\,{}{{x}^{2}}}}-{{4}\,{}{x}}}-{2}}+{8}}\\0 & = {{{-{{2}\,{}{{x}^{2}}}}-{{4}\,{}{x}}}+{6}}\\^{\text{divide each term by -1 to }}_{\text{make the first term positive}}\quad \rightarrow \quad0 &=2 x^{2} + 4 x - 6 \end{aligned}$

STEP: Solve the equation

0 = 2 x^{2} + 4 x - 6

$0 = 2 x^{2} + 4 x - 6$
[−2 points ⇒ 2 / 6 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = 2 x^{2} + 4 x - 6 \\ 0 & = (x + 3) (x - 1) \end{aligned}

$\begin{aligned} 0 & = 2 x^{2} + 4 x - 6 \\ 0 & = (x+3)(x-1) \end{aligned}$

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$ -values of our solutions. The $x$ -values are $-\text{3}$ and $\text{1}$ .

STEP: Calculate the corresponding values of

y

$y$
[−2 points ⇒ 0 / 6 points left]

By using the values of $x$ , we will calculate the corresponding values of $y$ . We will do so by substituting each $x$ -value into the linear equation.

NOTE: We can still get the same values of

y

$y$ if we substitute the values

x

$x$ into the equation

- 2 y = - 4 x - x y - 2

$- 2 y = - 4 x-x y - 2$ . If we do that, the calculations might become too long.

For $x = -3$ :

\begin{aligned} y & = 2 x + 4 \\ = - 2 \end{aligned}

$\begin{aligned} y &= 2 x + 4 \\ &= -2 \end{aligned}$

For $x = 1$ :

\begin{aligned} y & = 2 x + 4 \\ = 6 \end{aligned}

$\begin{aligned} y &= 2 x + 4 \\ &= 6 \end{aligned}$

The solutions are $x = -3$ together with $y = -2$ , and $x = 1$ with $y = 6$ . Using coordinate pairs, the solutions are $\left(-3; -2\right)$ and $\left(1; 6\right)$ .

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(-3; -2\right)$ and $\left(1; 6\right)$ . These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(-3; -2\right)$ and $\left(1; 6\right)$ .

Submit your answer as: and

Simultaneous equations with $xy$ terms

Solve for $x$ and $y$ from the given equations:

\begin{aligned} y & = - 2 x - x y + 1 \\ x & = 3 y + 5 \end{aligned}

$\begin{aligned} y & = - 2 x-x y + 1\\ x & = 3 y + 5 \end{aligned}$

INSTRUCTION:

You should get two coordinate pairs in the form $\left(x_1; y_1\right)$ and $\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting $3 y + 5$ in place of $x$ in the hyperbolic equation and then solve for $y$ .

STEP: Substitute the linear equation into the hyperbolic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

\begin{aligned} hyperbolic equation ⟶ & y = - 2 x - x y + 1 \\ linear equation ⟶ & x = 3 y + 5 \end{aligned}

$\begin{aligned} {\text{hyperbolic equation}}\quad \longrightarrow \quad & y = - 2 x-x y + 1\\ {\text{linear equation}}\quad \longrightarrow \quad & x = 3 y + 5 \end{aligned}$

This means we want two pairs of $x$ and $y$ -values which solve both equations.

We will substitute the value of $x$ from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

NOTE: We could also isolate either

x

$x$ or

y

$y$ from

y = - 2 x - x y + 1

$y = - 2 x-x y + 1$ and substitute its value into

x = 3 y + 5

$x = 3 y + 5$ . We could still get the same answer, but the calculations might become too long.

\begin{aligned} y & = - 2 (3 y + 5) - (3 y + 5) y + 1 \\ 0 & = - 2 (3 y + 5) - (3 y + 5) y + 1 - y \\ 0 & = - 2 (3 y + 5) - y (3 y + 5) + 1 - y \\ 0 & = - 3 y^{2} - 6 y - 5 y - y - 10 + 1 \\ 0 & = - 3 y^{2} - 11 y - y - 10 + 1 \\ 0 & = - 3 y^{2} - 12 y - 10 + 1 \\ 0 & = - 3 y^{2} - 12 y - 9 \\ _{make the first term positive}^{divide each term by -1 to} \to 0 & = 3 y^{2} + 12 y + 9 \end{aligned}

$\begin{aligned} y & = - 2 (3 y + 5)-(3 y + 5) y + 1\\ 0 & = - 2 (3 y + 5)-(3 y + 5) y + 1-y \\ 0 & = {{{{{-{2}}\,{}{\left({{{3}\,{}{y}}+{5}}\right)}}-{{y}\,{}{\left({{{3}\,{}{y}}+{5}}\right)}}}+{1}}-{y}}\\0 & = {{{{{{-{{3}\,{}{{y}^{2}}}}-{{6}\,{}{y}}}-{{5}\,{}{y}}}-{y}}-{10}}+{1}}\\0 & = {{{{{-{{3}\,{}{{y}^{2}}}}-{{11}\,{}{y}}}-{y}}-{10}}+{1}}\\0 & = {{{{-{{3}\,{}{{y}^{2}}}}-{{12}\,{}{y}}}-{10}}+{1}}\\0 & = {{{-{{3}\,{}{{y}^{2}}}}-{{12}\,{}{y}}}-{9}}\\^{\text{divide each term by -1 to }}_{\text{make the first term positive}}\quad \rightarrow \quad0 &=3 y^{2} + 12 y + 9 \end{aligned}$

STEP: Solve the equation

0 = 3 y^{2} + 12 y + 9

$0 = 3 y^{2} + 12 y + 9$
[−2 points ⇒ 2 / 6 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = 3 y^{2} + 12 y + 9 \\ 0 & = (y + 1) (y + 3) \end{aligned}

$\begin{aligned} 0 & = 3 y^{2} + 12 y + 9 \\ 0 & = (y+1)(y+3) \end{aligned}$

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $y$ -values of our solutions. The $y$ -values are $-\text{1}$ and $-\text{3}$ .

STEP: Calculate the corresponding values of

x

$x$
[−2 points ⇒ 0 / 6 points left]

By using the values of $y$ , we will calculate the corresponding values of $x$ . We will do so by substituting each $y$ -value into the linear equation.

NOTE: We can still get the same values of

x

$x$ if we substitute the values

y

$y$ into the equation

y = - 2 x - x y + 1

$y = - 2 x-x y + 1$ . If we do that, the calculations might become too long.

For $y = -1$ :

\begin{aligned} x & = 3 y + 5 \\ = 2 \end{aligned}

$\begin{aligned} x &= 3 y + 5 \\ &= 2 \end{aligned}$

For $y = -3$ :

\begin{aligned} x & = 3 y + 5 \\ = - 4 \end{aligned}

$\begin{aligned} x &= 3 y + 5 \\ &= -4 \end{aligned}$

The solutions are $x = 2$ together with $y = -1$ , and $x = -4$ with $y = -3$ . Using coordinate pairs, the solutions are $\left(2; -1\right)$ and $\left(-4; -3\right)$ .

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(2; -1\right)$ and $\left(-4; -3\right)$ . These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(2; -1\right)$ and $\left(-4; -3\right)$ .

Submit your answer as: and

Simultaneous equations with $xy$ terms

Solve for $x$ and $y$ :

\begin{aligned} - 2 y & = 6 x - x y + 2 \\ y & = 2 x - 10 \end{aligned}

$\begin{aligned} - 2 y & = 6 x-x y + 2\\ y & = 2 x - 10 \end{aligned}$

INSTRUCTION:

You should get two coordinate pairs in the form $\left(x_1; y_1\right)$ and $\left(x_2; y_2\right)$ . You must enter each pair of coordinates in one of the input boxes provided.
It does not matter which coordinate pair you type first.

Answer:

The solutions are and .

coordinate

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You need to start by substituting $2 x - 10$ in place of $y$ in the hyperbolic equation and then solve for $x$ .

STEP: Substitute the linear equation into the hyperbolic equation
[−2 points ⇒ 4 / 6 points left]

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

\begin{aligned} hyperbolic equation ⟶ & - 2 y = 6 x - x y + 2 \\ linear equation ⟶ & y = 2 x - 10 \end{aligned}

$\begin{aligned} {\text{hyperbolic equation}}\quad \longrightarrow \quad & - 2 y = 6 x-x y + 2\\ {\text{linear equation}}\quad \longrightarrow \quad & y = 2 x - 10 \end{aligned}$

This means we want two pairs of $x$ and $y$ -values which solve both equations.

We will substitute the value of $y$ from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

NOTE: We could also isolate either

x

$x$ or

y

$y$ from

- 2 y = 6 x - x y + 2

$- 2 y = 6 x-x y + 2$ and substitute its value into

y = 2 x - 10

$y = 2 x - 10$ . We could still get the same answer, but the calculations might become too long.

\begin{aligned} - 2 (2 x - 10) & = 6 x - x (2 x - 10) + 2 \\ 0 & = 6 x - x (2 x - 10) + 2 + 2 (2 x - 10) \\ 0 & = 6 x - (2 x^{2} - 10 x) + 2 + (4 x - 20) \\ 0 & = - 2 x^{2} + 16 x + 4 x + 2 - 20 \\ 0 & = - 2 x^{2} + 20 x + 2 - 20 \\ 0 & = - 2 x^{2} + 20 x - 18 \\ _{make the first term positive}^{divide each term by -1 to} \to 0 & = 2 x^{2} - 20 x + 18 \end{aligned}

$\begin{aligned} - 2 (2 x - 10) & = 6 x-x (2 x - 10) + 2\\ 0 & = 6 x-x (2 x - 10) + 2+2 (2 x - 10) \\ 0 & = {{{{{6}\,{}{x}}-{\left({{{2}\,{}{{x}^{2}}}-{{10}\,{}{x}}}\right)}}+{2}}+{\left({{{4}\,{}{x}}-{20}}\right)}}\\0 & = {{{{{-{{2}\,{}{{x}^{2}}}}+{{16}\,{}{x}}}+{{4}\,{}{x}}}+{2}}-{20}}\\0 & = {{{{-{{2}\,{}{{x}^{2}}}}+{{20}\,{}{x}}}+{2}}-{20}}\\0 & = {{{-{{2}\,{}{{x}^{2}}}}+{{20}\,{}{x}}}-{18}}\\^{\text{divide each term by -1 to }}_{\text{make the first term positive}}\quad \rightarrow \quad0 &=2 x^{2} - 20 x + 18 \end{aligned}$

STEP: Solve the equation

0 = 2 x^{2} - 20 x + 18

$0 = 2 x^{2} - 20 x + 18$
[−2 points ⇒ 2 / 6 points left]

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

NOTE: We could use the quadratic formula as well. In this case, it is not necessary to do so because we can factorise the equation.

\begin{aligned} 0 & = 2 x^{2} - 20 x + 18 \\ 0 & = (x - 1) (x - 9) \end{aligned}

$\begin{aligned} 0 & = 2 x^{2} - 20 x + 18 \\ 0 & = (x-1)(x-9) \end{aligned}$

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$ -values of our solutions. The $x$ -values are $\text{1}$ and $\text{9}$ .

STEP: Calculate the corresponding values of

y

$y$
[−2 points ⇒ 0 / 6 points left]

By using the values of $x$ , we will calculate the corresponding values of $y$ . We will do so by substituting each $x$ -value into the linear equation.

NOTE: We can still get the same values of

y

$y$ if we substitute the values

x

$x$ into the equation

- 2 y = 6 x - x y + 2

$- 2 y = 6 x-x y + 2$ . If we do that, the calculations might become too long.

For $x = 1$ :

\begin{aligned} y & = 2 x - 10 \\ = - 8 \end{aligned}

$\begin{aligned} y &= 2 x - 10 \\ &= -8 \end{aligned}$

For $x = 9$ :

\begin{aligned} y & = 2 x - 10 \\ = 8 \end{aligned}

$\begin{aligned} y &= 2 x - 10 \\ &= 8 \end{aligned}$

The solutions are $x = 1$ together with $y = -8$ , and $x = 9$ with $y = 8$ . Using coordinate pairs, the solutions are $\left(1; -8\right)$ and $\left(9; 8\right)$ .

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(1; -8\right)$ and $\left(9; 8\right)$ . These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(1; -8\right)$ and $\left(9; 8\right)$ .

Submit your answer as: and

3. Practical applications

Setting up simultaneous equations

Here are facts about two numbers:

The sum of the numbers is $\text{13}$ .
The numbers are consecutive numbers.

To find these numbers, we can write equations for each of these facts. Let $n_{1}$ represent one of the numbers and $n_{2}$ represent the other number. Then which equations accurately represent each fact? Select your answer from the choices below.

Answer:

Fact about the numbers	Equation
The sum of the numbers is $\text{13}$ .
The numbers are consecutive numbers.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.

STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

Word	Meaning
sum	+
product	$\times$
is	=
consecutive	1 apart
more than	add to
less than	subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let $n_{1}$ represent one of the numbers and $n_{2}$ represent the other number". So we can break up the first fact like this:

\begin{array}{ccc} \underset{⏟}{The sum of the numbers} & is & \underset{⏟}{13} \\ n_{1} + n_{2} & = & 13 \end{array}

$\begin{array}{ccc} \underbrace{\text{The sum of the numbers}} & \text{is} & \underbrace{\quad\text{13}\quad} \\ n_{1} + n_{2} & = & 13 \end{array}$

The correct equation for the first fact is $n_{1} + n_{2} = 13$ .

STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like $\text{10}$ and $\text{11}$ , or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is $\text{1}$ more than the small number." Another is "the small number is $\text{1}$ less than the larger number." Let's use the second option.

\begin{array}{ccc} \underset{⏟}{The smaller number} & is & \underset{⏟}{1 less than the larger number} \\ n_{1} & = & n_{2} - 1 \end{array}

$\begin{array}{ccc} \underbrace{\text{The smaller number}} & \text{is} & \underbrace{\text{1 less than the larger number}} \\ n_{1} & = & n_{2} - 1 \end{array}$

This makes sense: consecutive numbers are separated by $\text{1}$ . The equation shows that with the $-\text{1}$ .

The correct answers are:

Fact about the numbers	Equation
The sum of the numbers is $\text{13}$ .	$n_1 + n_2 = 13$
The numbers are consecutive numbers.	$n_{1} = n_{2} - 1$

Submit your answer as: and

Word problems: distance, speed and time

Two ships start out at two different ports. The ships are $\text{141}\ \text{km}$ apart. The ships start moving towards each other. One ship is moving at $\text{72}\ \text{km/h}$ and the other ship at $\text{69}\ \text{km/h}$ .

If both ships started their journey at the same time, how long will they take to pass each other?

Answer: The two ships pass each other after hours.

numeric

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.

STEP: Sketch a simple picture to organise the information
[−2 points ⇒ 4 / 6 points left]

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two ships must be equal to the total distance when the ships meet:

\begin{aligned} d_{1} + d_{2} & = d_{t o t a l} \\ ↓ \\ d_{1} + d_{2} & = 141 km \end{aligned}

$\begin{aligned} d_{1} + d_{2} & = d_{total} \\ \quad & \downarrow \quad \\ d_{1} + d_{2} & = \text{141}\ \text{km} \end{aligned}$

STEP: Write equations to describe the motion of the ships
[−2 points ⇒ 2 / 6 points left]

This question is about distances, speeds, and times. The equation connecting these values is

\begin{aligned} speed & = \frac{distance}{time} \\ OR \\ distance & = speed \times time \end{aligned}

$\begin{aligned} \text{speed } & = \frac{\text{distance }}{\text{time}} \\ \quad & \text{ OR } \quad \\ \text{distance } & = \text{speed } \times \text{time} \end{aligned}$

We want to know the amount of time needed for the ships to meet. Let the time taken be $t$ . Then we can write an expression for the distance each of the ships travels:

For ship 1:

\begin{aligned} d_{1} & = s_{1} t \\ _{ship is 72 km/h}^{The speed of the first} \to d_{1} & = 72 t \end{aligned}

$\begin{aligned} d_{1} &= s_{1} t \\ ^{\text{The speed of the first}}_{\text{ship is } \text{72}\ \text{km/h}} \quad \rightarrow \quad d_{1} &= \text{72}t \end{aligned}$

For ship 2:

\begin{aligned} d_{2} & = s_{2} t \\ _{ship is 69 km/h}^{The speed of the second} \to d_{2} & = 69 t \end{aligned}

$\begin{aligned} d_{2} &= s_{2} t \\ ^{\text{The speed of the second}}_{\text{ship is } \text{69}\ \text{km/h}} \quad \rightarrow \quad d_{2} &= \text{69}t \end{aligned}$

STEP: Solve the equations simultaneously for

t

$t$
[−2 points ⇒ 0 / 6 points left]

Now we have three different equations. We can combine them using substitution to solve for the value of $t$ .

\begin{aligned} d_{1} + d_{2} & = 141 \\ (72 t) + (69 t) & = 141 \\ 141 t & = 141 \\ t & = \frac{141}{141} \\ t & = 1 \end{aligned}

$\begin{aligned} d_{1} + d_{2} &= \text{141} \\ (\text{72}t) + (\text{69}t) &= \text{141} \\ \text{141}t &= \text{141} \\ t &= \frac{\text{141}}{\text{141}} \\ t &= \text{1} \end{aligned}$

The ships will meet after $\text{1}$ hours.

Submit your answer as:

Word problems: solving simultaneous equations

A group of friends is buying lunch together. The group buys $\text{2}$ wraps and $\text{4}$ hamburgers. Here are some facts about their lunch:

the total cost for the $\text{2}$ wraps and $\text{4}$ hamburgers is $\text{R}\,\text{166}$
a wrap costs $\text{R}\,\text{8}$ more than a hamburger

What is the price for one wrap and the price for one hamburger?

Answer:

A wrap costs $\text{R}$ and a hamburger costs $\text{R}$ .

numeric

HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

You need to choose variables to represent the things you want to find and write equations based on the information in the question.

STEP: Pick variables for the things we want to know
[−1 point ⇒ 6 / 7 points left]

The two things we want to know in this question are the prices for each of the items (a wrap and a hamburger). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

\begin{aligned} w & = the price of a wrap \\ h & = the price of a hamburger \end{aligned}

$\begin{aligned} w & = \text{the price of a wrap} \\ h & = \text{the price of a hamburger} \end{aligned}$

STEP: Write equations based on the information in the question
[−2 points ⇒ 4 / 7 points left]

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the $\text{2}$ wraps and $\text{4}$ hamburgers is $\text{R}\,\text{166}$ ." We can use the expression $2 w$ to represent the price of the $\text{2}$ wraps. Similarly, the expression $4 h$ represents the price of the $\text{4}$ hamburgers. With these values we can write a full equation for the prices:

\begin{aligned} In words: & the total cost for the 2 wraps and 4 hamburgers is R 166 \\ In maths: & 2 w + 4 h = 166 \end{aligned}

$\begin{aligned} \text{In words: } & \;\; \text{the total cost for the 2 wraps and 4 hamburgers is } \text{R}\,\text{166} \\ \text{In maths: } & \;\; 2 w + 4 h = 166 \end{aligned}$

Now we can use the second point. It says, "a wrap costs $\text{R}\,\text{8}$ more than a hamburger." We can write this as an equation like this:

\begin{aligned} In words: & a wrap costs R 8 more than a hamburger \\ In maths: & w = h + 8 \end{aligned}

$\begin{aligned} \text{In words: } & \;\; \text{a wrap costs } \text{R}\,\text{8} \text{ more than a hamburger} \\ \text{In maths: } & \;\; w = h+8 \end{aligned}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 2 / 7 points left]

Now that we have two equations, we need to solve them simultaneously.

\begin{aligned} 2 w + 4 h & = 166 \\ w & = h + 8 \end{aligned}

$\begin{aligned} 2 w + 4 h & = 166 \\ w & = h+8 \end{aligned}$

We could use elimination, but substitution is a better choice (because $w$ is already the subject). Substitute the second equation into the first equation and solve!

\begin{aligned} 2 w + 4 h & = 166 \\ 2 (h + 8) + 4 h & = 166 \\ 2 h + 16 + 4 h & = 166 \\ 6 h & = 166 - 16 \\ h & = \frac{150}{6} \\ = 25 \end{aligned}

$\begin{aligned} 2 w + 4 h &= 166 \\ 2 \left( h+8 \right) +4 h &= 166 \\ 2 h + 16 + 4 h &= 166 \\ 6 h &= 166 - 16 \\ h &= \frac{150} {6} \\ &= 25 \end{aligned}$

This means that the price of one hamburger is $\text{R}\,\text{25}$ .

STEP: Find the other variable's value
[−1 point ⇒ 1 / 7 points left]

Finally, use the value we found for $h$ to find the value of $w$ (the price of a wrap). We can use either equation to do this, but the second one is easier to use (because $w$ is already isolated).

\begin{aligned} w & = h + 8 \\ = 25 + 8 \\ = 33 \end{aligned}

$\begin{aligned} w &= h+8 \\ &= 25+8 \\ &= 33 \end{aligned}$

The price for one wrap is $\text{R}\,\text{33}$ .

STEP: Write the final answer
[−1 point ⇒ 0 / 7 points left]

It is important to write the answer to a word problem as a complete sentence.

The price of the wrap is $\text{R}\,\text{33}$ while a hamburger costs $\text{R}\,\text{25}$ .

Submit your answer as: and

Word problems: checking answers about shopping

Suppose you must solve this word problem:

At a shop there are some fruits and sweets for sale. Ben buys some bananas and some sweets. The total cost for $\text{5}$ bananas and $\text{5}$ sweets is $\text{R}\,\text{32,50}$ . And each sweet costs $\text{R}\,\text{0,30}$ less than each banana. What is the price for one banana?

The answer to this question is a number. Answer the two questions below about this number.

Answer:

Can the price be a decimal number (or must it be an integer)?
Can the price be a negative number?

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the price of each banana. Can the price be a decimal, like $\text{4,5}$ ? Can it be negative, like $-\text{5}$ ?

STEP: Decide if the price can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be $\text{R}\,\text{6,75}$ . Or $\text{R}\,\text{5}$ . Or $\text{R}\,\text{19,87}$ . The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the price can be a decimal number or not. In other words, can the price of something be a decimal number like $\text{R}\,\text{4,30}$ , or must a price always be a whole number like $\text{R}\,\text{4,00}$ ? Of course we know that many prices are not whole numbers! This is similar to the example above about airtime, which can be a decimal number. There is no reason why the price of an item must be a whole number.

The answer for the first question is: Yes, the price can be a decimal number.

STEP: Decide if the price can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if the price can be negative, or must be positive. Prices cannot be negative. Imagine something which costs - $\text{R}\,\text{5,50}$ . That is impossible!

The answer to the second question is: No, the price cannot be a negative number.

The correct answer choices are:

Yes, it can be a decimal
No, it cannot be negative

Submit your answer as: and

Word problems: checking answers with shapes

Suppose you must solve this word problem:

Yaseen is drawing a pattern of rectangles in his notebook. The number of rectangles in each figure is $\text{2}$ more than the number of rectangles in the previous figure. If there are $\text{10}$ rectangles in the fourth figure, how many rectangles are there in the first figure?

The answer to this question is a number. Answer the two questions below about this number.

Answer:

Can the number of rectangles be a decimal number (or must it be an integer)?
Can the number of rectangles be a negative number?

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the number of rectangles in the first figure of Yaseen's diagram. Can the number of rectangles be a decimal, like $\text{4,5}$ ? Can it be negative, like $-\text{5}$ ?

STEP: Decide if the number of rectangles can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be $-\text{150}$ days of school, shame. Similarly, it is not possible to have $\text{176,3562}$ days of school. What type of numbers are realistic for the number of rectangles?

For the first question we must decide if the number of rectangles can be a decimal number. Is that fine or is it impossible?

In other words, can the number of rectangles be a decimal number like $\text{3,7}$ , or must the number be a whole number like $\text{4}$ ? Just like the number of days in the school year, the number of rectangles cannot be a decimal value. $\text{3,7}$ rectangles does not make sense! The number of rectangles must be a whole number! (You might think, "That's wrong, we can just draw part of a rectangle to get $\text{3,7}$ rectangles." But if the shape is not complete, it is not a rectangle.)

The answer for the first question is: No, it must be an integer, the number of rectangles cannot be a decimal number.

STEP: Decide if the number of rectangles can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if number of rectangles can be negative, or must be positive. The number of rectangles cannot be negative. If Yaseen drew any rectangles at all, then the answer to this question must be positive. There is no such thing as $-\text{4}$ rectangles.

The answer for the second question is: No, the number of rectangles cannot be a negative number.

The correct answer choices are:

No, it must be an integer
No, it cannot be negative

Submit your answer as: and

Word problems: test scores and simultaneous equations

Hassana and Chibueze are friends. Hassana takes Chibueze's consumer studies test paper and says: “I have $\text{20}$ marks more than you do and the sum of both our marks is equal to $\text{134}$ . What are our marks?”

Answer:

Hassana got marks and Chibueze got marks for the consumer studies test paper.

numeric

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.

STEP: Pick variables for the marks
[−1 point ⇒ 5 / 6 points left]

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

\begin{aligned} h & = Hassana's mark \\ c & = Chibueze's mark \end{aligned}

$\begin{aligned} h &= \text{ Hassana's mark} \\ c &= \text{ Chibueze's mark} \end{aligned}$

STEP: Write equations about the students' marks
[−2 points ⇒ 3 / 6 points left]

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

\begin{aligned} _{more than Chibueze}^{Hassana has 20 marks} & \to h = c + 20 \\ _{(the total) is 134}^{The sum of the marks} & \to h + c = 134 \end{aligned}

$\begin{aligned} ^{\text{Hassana has } \text{20} \text{ marks}} _{\text{ more than Chibueze}} \quad & \rightarrow \quad h = c + \text{20} \\ \\ ^{\text{The sum of the marks}} _{\text{(the total) is } \text{134}} \quad & \rightarrow \quad h+c = \text{134} \end{aligned}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

\begin{aligned} h + c & = 134 \\ (c + 20) + c & = 134 \\ 2 c & = 134 - 20 \\ c & = \frac{114}{2} \\ = 57 \end{aligned}

$\begin{aligned} h+c &= \text{134} \\ (c+\text{20})+c &= \text{134} \\ 2c &= 134 - \text{20}\\ c &= \frac{\text{114}}{\text{2}}\\ &= \text{57} \end{aligned}$

This means that Chibueze's mark is $\text{57}$ .

STEP: Find Hassana's mark
[−1 point ⇒ 0 / 6 points left]

Now we can use Chibueze's mark and one of the equations we have to find Hassana's mark. The easiest way to do that is to substitute Chibueze's mark back into the first equation:

\begin{aligned} h & = c + 20 \\ = (57) + 20 \\ = 77 \end{aligned}

$\begin{aligned} h &= c+\text{20} \\ &= \left( \text{57} \right) + \text{20} \\ &= \text{77} \end{aligned}$

The students achieved these marks: Hassana earned $\text{77}$ and Chibueze earned $\text{57}$ .

Submit your answer as: and

Word problems: odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

The numbers are consecutive odd integers.
The sum of the numbers is $\text{24}$ .

Determine the value of the larger number.

Answer: The larger number is .

numeric

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by choosing variables for the two numbers you are trying to find. For example, you can use

n_{1}

$n_{1}$ and

n_{2}

$n_{2}$ . Then write equations using these variables based on the facts given in the question.

STEP: Pick variables
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find one of them (the larger one). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a sum of $\text{24}$ . We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

\begin{aligned} n_{1} & = the smaller number \\ _{we need to find}^{this is the number} \to n_{2} & = the larger number \end{aligned}

$\begin{aligned} n_{1} &= \text{the smaller number} \\ ^{\text{this is the number}}_{\text{we need to find}} \quad \rightarrow \quad n_{2} &= \text{the larger number} \end{aligned}$

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact given in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, $\text{11}$ and $\text{13}$ are consecutive odd numbers. Since we defined $n_{1}$ as the smaller number, $n_{2}$ must be $\text{2}$ more than $n_{1}$ .

\begin{matrix} n_{2} = n_{1} + 2 \end{matrix}

$\begin{array}{c}n_{2} = n_{1} + 2\end{array}$

The " $\text{+2}$ " skips the even integer which sits between $n_{1}$ and $n_{2}$ .

The second fact about the numbers tells us that "the sum of the numbers is $\text{24}$ ". Remember that sum means addition. So:

\begin{matrix} n_{1} + n_{2} = 24 \end{matrix}

$\begin{array}{c}n_{1} + n_{2} = 24\end{array}$

STEP: Solve the equations for

n_{2}

$n_{2}$
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the larger number (we do not need both of them). That means we need to find the value of $n_{2}$ .

We can solve this using substitution. However, remember that we want the value of $n_2$ (the larger number). We can start by rearranging the first equation to make $n_1$ the subject. Then the substitution step will remove $n_1$ from the second equation and we can solve for $n_2$ . (This is not required - it just makes the solution faster.)

\begin{aligned} n_{2} & = n_{1} + 2 \\ n_{2} - 2 & = n_{1} \end{aligned}

$\begin{aligned} n_2 &= n_1 + 2 \\ n_2 - 2 &= n_1 \end{aligned}$

Now substitute $n_2 - 2$ into the other equation and solve for $n_2$ .

\begin{aligned} (n_{2} - 2) + n_{2} & = 24 \\ 2 n_{2} - 2 & = 24 \\ 2 n_{2} & = 26 \\ n_{2} & = 13 \end{aligned}

$\begin{aligned} \left(n_2 - 2\right) + n_2 &= 24 \\ 2n_2 - 2 &= 24 \\ 2n_2 &= 26 \\ n_2 &= 13 \end{aligned}$

The result is $n_{2} = 13$ . Notice that this means that the other number, $n_{1}$ , must be $\text{11}$ , because $n_{2} = n_{1} + 2$ . This is perfect, because we know that the sum of the numbers is $\text{24}$ , and $13 + 11 = 24$ .

The larger number is $\text{13}$ .

Submit your answer as:

Word problems: checking answers

Emma is $\text{12}$ years younger than her brother, Songezo. In $\text{10}$ years, Songezo will be $\text{2}$ times as old as Emma. How old is Emma now?

INSTRUCTION: If there is no acceptable solution, type 'no solution' in the answer box.

Answer:

Emma is years old.

numeric

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to find Emma's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.

STEP: Choose variables for each person's age
[−1 point ⇒ 4 / 5 points left]

We need to determine Emma's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Songezo and Emma. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

\begin{aligned} a_{S} & = Songezo's age \\ a_{E} & = Emma's age \end{aligned}

$\begin{aligned} a_{S} & = \text{Songezo's age} \\ a_{E} & = \text{Emma's age} \end{aligned}$

STEP: Write an equation based on the information in the question
[−1 point ⇒ 3 / 5 points left]

Now let's write equations using these variables. From the question we know that: "Emma is $\text{12}$ years younger than her brother, Songezo." We need to translate that into mathematics. The key word is younger, which tells us to use subtraction to relate the ages.

\begin{aligned} a_{E} = a_{S} - 12 \end{aligned}

$\begin{aligned} a_E = a_S - 12 \end{aligned}$

STEP: Write the equations
[−1 point ⇒ 2 / 5 points left]

We also know that: "In $\text{10}$ years, Songezo will be $\text{2}$ times as old as Emma." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

\begin{aligned} a_{S} + 10 & = Songezo's age in 10 years \\ a_{E} + 10 & = Emma's age in 10 years \end{aligned}

$\begin{aligned} a_{S} + 10 & = \text{Songezo's age in 10 years} \\ a_{E} + 10 & = \text{Emma's age in 10 years} \end{aligned}$

These are the ages at which Songezo will be $\text{2}$ times as old as Emma. We can put all this together as follows:

\begin{array}{rcl} _{in 10 years}^{Songezo's age} & = & 2 \times_{in 10 years}^{Emma's age} \\ ↓ \\ a_{S} + 10 & = & 2 (a_{E} + 10) \end{array}

$\begin{array}{rcl} ^{\text{Songezo's age}}_{\text{in 10 years}} & = & 2\; \times\; ^{\text{Emma's age}}_{\text{in 10 years}} \\ \quad & \downarrow & \quad \\ a_{S} + 10 & = & 2 \left( a_{E} + 10 \right) \end{array}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 0 / 5 points left]

Now we have to solve two equations, both of them including the variables $a_{S}$ and $a_{E}$ . The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is $a_E = a_S - 12$ . We can substitute this into the second equation and solve for $a_{S}$ .

\begin{aligned} a_{S} + 10 & = 2 (a_{E} + 10) \\ a_{S} + 10 & = 2 ((a_{S} - 12) + 10) \\ a_{S} + 10 & = 2 a_{S} - 4 \\ 14 & = a_{S} \end{aligned}

$\begin{aligned} a_{S} + 10 &= 2 \left(a_{E} + 10\right) \\ a_{S} + 10 &= 2 \left(\left(a_{S} - 12\right) + 10\right) \\ a_{S} + 10 &= 2 a_{S} - 4 \\ 14 &= a_{S} \\ \quad \end{aligned}$

Terrific: this means that Songezo is $\text{14}$ years old. But the question asked for us to find Emma's age. We can find it using the first equation, which relates the two ages:

\begin{aligned} a_{E} & = a_{S} - 12 \\ = (14) - 12 \\ = 2 \end{aligned}

$\begin{aligned} a_E & = a_S - 12 \\ & = \left(14\right) - 12 \\ & = 2 \end{aligned}$

So we finally got the answer to the question. Emma is $\text{2}$ years old.

The correct answer is: 2.

Submit your answer as:

Word problems: products of odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

The numbers are consecutive even integers.
The product of the numbers is $\text{48}$ .

Find the value of the smaller number.

Answer: The smaller number is .

numeric

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by choosing variables for the two numbers you are trying to find. Then write equations using these variables based on the information given in the question.

STEP: Pick variables for the numbers
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find the smaller number (not both). We know certain things about these numbers: they are positive, they are consecutive even integers, and they have a product of $\text{48}$ . We can use that information to solve this question using simultaneous equations.

The first thing we need to do is define variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent the things we want to find. In this case we are looking for two numbers, so these are good choices:

\begin{aligned} _{we need to find}^{this is the number} \to n_{1} & = the smaller number \\ n_{2} & = the larger number \end{aligned}

$\begin{aligned} ^{\text{this is the number}}_{\text{we need to find}} \quad \rightarrow \quad n_{1} &= \text{the smaller number} \\ n_{2} &= \text{the larger number} \end{aligned}$

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact in the question says that the numbers are "consecutive even integers". So both of the numbers are even, and they come one after another. For example, $\text{10}$ and $\text{12}$ are consecutive even numbers. Since we defined $n_{1}$ as the smaller number, $n_{2}$ must be $\text{2}$ more than $n_{1}$ .

\begin{matrix} n_{2} = n_{1} + 2 \end{matrix}

$\begin{array}{c}n_{2} = n_{1} + 2\end{array}$

The " $\text{+2}$ " skips the odd integer which sits between $n_{1}$ and $n_{2}$ .

The second fact about the numbers tells us that "the product of the numbers is $\text{48}$ ". Remember that product means multiplication. So:

\begin{matrix} n_{1} \cdot n_{2} = 48 \end{matrix}

$\begin{array}{c}n_{1} \cdot n_{2} = 48\end{array}$

STEP: Solve the equations for

n_{1}

$n_{1}$
[−2 points ⇒ 0 / 5 points left]

We can now use these two equations to find the answer to the question. But remember that we only need to find the smaller number (we do not need both of them). That means we need to find the value of $n_{1}$ .

If we subsitute $n_2 = n_1 + 2$ into the equation $n_1 \cdot n_2 = 48$ , the $n_2$ terms will disappear, leaving $n_1$ . Then we can solve for $n_1$ , which is exactly what we want.

\begin{aligned} n_{1} \cdot (n_{1} + 2) & = 48 \\ n_{1}^{2} + 2 n_{1} & = 48 \\ n_{1}^{2} + 2 n_{1} - 48 & = 0 \\ (n_{1} + 8) (n_{1} - 6) & = 0 \end{aligned}

$\begin{aligned} n_1 \cdot \left(n_1 + 2\right) &= 48 \\ n_1^{\; 2} + 2n_1 &= 48 \\ n_1^{\; 2} + 2n_1 - 48 &= 0 \\ \left(n_1 + 8\right) \left(n_1 - 6\right) &= 0 \end{aligned}$

\begin{aligned} n_{1} = - 8 and n_{1} = 6 \end{aligned}

$\begin{aligned} n_1 = -8 \quad \text{and} \quad n_1 = 6 \end{aligned}$

This solution led to two answers for $n_{1}$ . However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves $n_{1} = 6$ .

This means the other number, $n_{2}$ , must be $\text{8}$ , because $n_{2} = n_{1} + 2$ . This is perfect, because we also know that the product of the numbers is $\text{48}$ , and $6 \cdot 8 = 48$ .

The smaller number is $\text{6}$ .

Submit your answer as:

Word problems: finding consecutive numbers

Here are some facts about two numbers:

the numbers are consecutive integers
both numbers are positive
the product of the numbers is $\text{210}$

What are the two numbers?

TIP: 'Product' means multiplication; 'consecutive' means 'following each other without gaps between.'

INSTRUCTION: Write your answers in the boxes below. It does not matter what order your answers are in.

Answer: The numbers are and .

numeric

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You should represent the numbers in the question by $n$ and $n+1$ . Write an equation based on the fact that the product of those two numbers is $\text{210}$ .

STEP: Pick variables to represent the numbers
[−1 point ⇒ 5 / 6 points left]

This question asks us to find two numbers which have a product of $\text{210}$ . That means if we multiply the numbers we get $\text{210}$ .

The first thing we need to do is pick variables for the numbers we are trying to find. We can pick any variable. But it is a good idea to pick a variable which is related to whatever it represents. In this question, $n$ is a good choice because we are looking for numbers. In fact, since there are two different numbers, we can do this:

\begin{aligned} n_{1} & = the first number \\ n_{2} & = the next number \end{aligned}

$\begin{aligned} n_{1} & = \text{the first number} \\ n_{2} & = \text{the next number} \end{aligned}$

STEP: Write an equation based on the information in the question
[−1 point ⇒ 4 / 6 points left]

The question tells us that the product of these numbers is $\text{210}$ . In other words, if we multiply the numbers, we get $\text{210}$ . As an equation this is:

\begin{matrix} n_{1} \cdot n_{2} = 210 \end{matrix}

$\begin{array}{c}n_{1} \cdot n_{2} = 210\end{array}$

STEP: Connect the two variables
[−1 point ⇒ 3 / 6 points left]

Now we can use the fact that the numbers are consecutive. So if $n_{1}$ represents the first number, then the next number must be one more than $n_{1}$ .

\begin{aligned} If: & n_{2} = the number following n_{1} \\ then: & n_{2} = n_{1} + 1 \end{aligned}

$\begin{aligned} \text{If:} \qquad & n_{2} = \text{the number following } n_{1} \\ \\ \text{then:} \qquad & n_{2} = n_{1} + 1 \end{aligned}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have two equations about the two numbers.

\begin{aligned} n_{1} \cdot n_{2} = 210 \\ n_{2} = n_{1} + 1 \end{aligned}

$\begin{aligned} & n_{1} \cdot n_{2} = 210 \\ & n_{2} = n_{1} + 1 \end{aligned}$

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for $n_{1}$ .

\begin{aligned} n_{1} \cdot (n_{1} + 1) & = 210 \\ (n_{1})^{2} + n_{1} - 210 & = 0 \\ (n_{1} + 15) (n_{1} - 14) & = 0 \end{aligned}

$\begin{aligned} n_{1} \cdot \left(n_{1} + 1 \right) & = 210 \\ (n_{1})^2 + n_{1} - 210 &= 0 \\ \left( n_{1} + 15\right) \left(n_{1}-14 \right) &= 0 \end{aligned}$

\begin{aligned} n_{1} = - 15 and n_{1} = 14 \end{aligned}

$\begin{aligned} n_{1} = -15 \quad \text{and} \quad n_{1} = 14 \end{aligned}$

The solutions to the equation are $n_{1} = -15$ or $n_{1} = 14$ . This means the first number is either $\text{−15}$ or $\text{14}$ .

STEP: Find the final answers
[−1 point ⇒ 0 / 6 points left]

Wait a minute! The numbers $\text{−15}$ and $\text{14}$ are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are positive. That means we must throw out the $n_{1} = -15$ , so the first number is $n_{1} = 14$ .

Based on that we can find the second number. The second number is $n_{1} + 1$ , which is equal to $\left(14\right) + 1 = 15$ .

The two consecutive integers are $\text{14}$ and $\text{15}$ .

Submit your answer as: and

Word problems: an age-old question

Talwar has a son, Roy. Here are some facts about how old Talwar and Roy are:

Talwar is $\text{9}$ times as old as Roy right now.
$\text{9}$ years from now, Talwar will be $\text{3}$ times as old as Roy.

How old are Talwar and Roy now?

Answer:

Talwar is years old and Roy is years old.

numeric

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.

STEP: Choose variables for the information
[−1 point ⇒ 5 / 6 points left]

In this question we want to find the age of two people: Talwar and his son, Roy. We can solve this by setting up two equations and solving them simultaneously.

Start by choosing variables to represent the ages of the father and the son. (We need to do this because we don't know the ages!) It is a good idea to choose variables that match what we are describing. So:

\begin{aligned} Let t = Talwar's age \\ Let r = Roy's age \end{aligned}

$\begin{aligned} & \text{Let } t = \text{Talwar's age} \\ & \text{Let } r = \text{Roy's age} \end{aligned}$

STEP: Write equations based on the information in the question
[−2 points ⇒ 3 / 6 points left]

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Talwar is now $\text{9}$ times as old as Roy." As an equation, this is:

\begin{matrix} Ages now: t = 9 r \end{matrix}

$\begin{array}{c}\text{Ages now: } \quad t = \text{9} r\end{array}$

The second piece of information says that in " $\text{9}$ years... Talwar will be $\text{3}$ times as old as his son." In $\text{9}$ years Talwar will be $t + \text{9}$ years old, and similarly Roy will be $r + \text{9}$ years old. Then:

\begin{matrix} Ages in 9 years: t + 9 = 3 (r + 9) \end{matrix}

$\begin{array}{c}\text{Ages in } \text{9} \text{ years: } \quad t + \text{9} = \text{3}(r+\text{9})\end{array}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have a pair of simultaneous equations! Substitute the first equation into the second equation and solve. (You can solve the equations using elimination if you prefer. In that case, the easiest option is to subtract the equations to cancel $t$ .)

\begin{aligned} t + 9 & = 3 (r + 9) \\ (9 r) + 9 & = 3 r + 27 \\ 6 r & = 18 \\ r & = 3 \end{aligned}

$\begin{aligned} t + \text{9} &= \text{3}(r+\text{9}) \\ (\text{9}r) + \text{9} &= \text{3}r+\text{27} \\ 6 r &= \text{18} \\ r & = \text{3} \end{aligned}$

Great! Now we know that Roy is $\text{3}$ years old.

STEP: Use Roy's age to find Talwar's age
[−1 point ⇒ 0 / 6 points left]

We can now find Talwar's age. Substitute Roy's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

\begin{aligned} t & = 9 r \\ = 9 (3) \\ = 27 \end{aligned}

$\begin{aligned} t &= \text{9}r \\ &= \text{9} \left( \text{3} \right) \\ &= \text{27} \end{aligned}$

Write your final answer: Talwar is $\text{27}$ years old and Roy is $\text{3}$ years old.

Submit your answer as: and

Word Problems: checking answers

Suppose you must solve this word problem:

A marathon is a long-distance running race with an official distance of $\text{42,195}\ \text{kilometres}$ . $\text{1 000}$ people ran the "Sunshine and Roses are for Sissies" marathon in Durban. The second-place runner finished $\text{12,72}$ seconds after the winner. The third runner was $\text{13,55}$ seconds behind the second runner, and the fourth runner was $\text{26,15}$ seconds behind the third runner. If the second-place runner had an average speed of $\text{19,15}\ \text{km/h}$ , how many of these runners finished the marathon in less than $\text{2,24}$ hours?

Can the answer to the question above be a negative number?

Answer:

Can the answer be negative?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Think about the meaning of the number you need. Some values cannot be decimal numbers. Other values cannot be negative numbers. For example, a distance can be a decimal, but it cannot be negative.

STEP: Decide if the answer can be negative
[−1 point ⇒ 0 / 1 points left]

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a negative number or not.

Focus on what the problem asks: how many of these runners finished the marathon in less than $\text{2,24}$ hours. So the question is about the number of people who finished the race before $\text{2,24}$ hours had passed. The answer cannot be negative. The number of people to finish before $\text{2,24}$ hours might be $\text{3}$ or $\text{7}$ people, but it cannot be $-\text{5}$ people. In other words, the number of people must be a positive number or zero, but it cannot be negative.

The correct answer is: No, it cannot be negative.

Submit your answer as:

Word problems: rectangle facts

The diagonal of a rectangle is $\text{16} \ \text{cm}$ more than its width. The length of the same rectangle is $\text{8}\ \text{cm}$ more than its width.

Determine the width and length of the rectangle.

Answer: The width is

cm

$\text{cm}$ and length is

cm

$\text{cm}$ .

numeric

HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.

STEP: Draw a diagram and label it
[−1 point ⇒ 7 / 8 points left]

Drawing a diagram is a helpful way to organise the information in a question about a shape. This question is about a rectangle, so we can start with a rectangle picture. The question mentions the diagonal of the ractangle, so we should draw that also.

Notice we have added labels for the three parts of the rectangle mentioned in the question: $d$ is the diagonal, $w$ is the width, and $l$ is the length.

STEP: Connect the variables in the diagram
[−2 points ⇒ 5 / 8 points left]

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

\begin{array}{lcl} _{16 cm more than its width}^{the diagonal of the rectangle is} & \to d & = w + 16 \\ _{8 cm more than its width}^{the length of the rectangle is} & \to l & = w + 8 \end{array}

$\begin{array}{lcl} ^{\text{the diagonal of the rectangle is}}_{ \text{16} \ \text{cm} \text{ more than its width}} & \rightarrow \quad d &= w + \text{16} \\ \\ ^{\text{the length of the rectangle is}}_{\text{8}\ \text{cm} \text{ more than its width}} & \rightarrow \quad l &= w + \text{8} \end{array}$

STEP: Use the Theorem of Pythagoras and solve the equations simultaneously
[−2 points ⇒ 3 / 8 points left]

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

From the picture we can see that there are two right-angled triangles in the rectangle. As always for a right-angled triangle, we can use the Theorem of Pythagoras. Using our variables from the diagram, we can write:

\begin{aligned} d^{2} & = w^{2} + l^{2} \end{aligned}

$\begin{aligned} d^2 &= w^2 + l^2 \end{aligned}$

And now we can substitute in the equations from above. Specifically we can substitute $w + \text{16}$ in for $d$ and $w + \text{8}$ in for $l$ . Then expand the binomials and simplify the equation as much as possible.

\begin{aligned} d^{2} & = w^{2} + l^{2} \\ (w + 16)^{2} & = w^{2} + (w + 8)^{2} \\ w^{2} + 32 w + 256 & = w^{2} + (w^{2} + 16 w + 64) \\ 0 & = w^{2} - 16 w - 192 \end{aligned}

$\begin{aligned} d^2 &= w^2 + l^2 \\ (w + \text{16})^2 & = w^2 + (w + \text{8})^2\\ w^2 + \text{32}w + \text{256} & = w^2 + (w^2 + \text{16}w + \text{64})\\ 0 & = w^2 - \text{16}w - \text{192} \end{aligned}$

STEP: Solve the equation for

w

$w$
[−2 points ⇒ 1 / 8 points left]

We have a quadratic equation in standard form! It is time to solve it.

\begin{aligned} 0 & = w^{2} - 16 w - 192 \\ 0 & = (w - 24) (w + 8) \\ w & = 24 or w = - 8 \end{aligned}

$\begin{aligned} 0 & = w^2 - \text{16}w - \text{192}\\ 0 & = (w -\text{24})(w + \text{8})\\ w & = \text{24} \quad \text{or } \quad w = -\text{8} \end{aligned}$

This means that the width of the rectangle is either $\text{24} \ \text{cm}$ or $-\text{8} \ \text{cm}$ . But the dimensions of a rectangle cannot be negative, so $w = \text{24} \ \text{cm}$ .

STEP: Calculate the length of the rectangle
[−1 point ⇒ 0 / 8 points left]

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects $l$ to $w$ .

\begin{aligned} l & = w + 8 \\ = (24) + 8 \\ = 32 cm \end{aligned}

$\begin{aligned} l &= w + \text{8} \\ &= \left(\text{24}\right) + \text{8} \\ & = \text{32}\ \text{cm} \end{aligned}$

The width of the rectangle is $\text{24} \ \text{cm}$ and the length is $\text{32} \ \text{cm}$ .

Submit your answer as: and

Setting up simultaneous equations

Last week, Anthea and Bokamoso had a physics test. Now they are comparing their marks and they notice these facts:

The sum of the marks is $\text{160}$ .
Anthea's mark is $\text{8}$ less than Bokamoso's mark.

Let $a$ represent Anthea's mark and $b$ represent Bokamoso's mark. Then which equations below accurately represent the facts? Select your answer from the choices below.

Answer:

Fact about the test scores	Equation
The sum of the marks is $\text{160}$ .
Anthea's mark is $\text{8}$ less than Bokamoso's mark.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is less. Use these key words to change the words into operations, for example, addition or multiplication.

STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

Word	Meaning
sum	+
product	$\times$
is	=
consecutive	1 apart
more than	add to
less than	subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable $a$ for Anthea's mark and $b$ for Bokamoso's mark. So we can break up the first fact like this:

\begin{array}{ccc} \underset{⏟}{The sum of the marks} & is & \underset{⏟}{160} \\ a + b & = & 160 \end{array}

$\begin{array}{ccc} \underbrace{\text{The sum of the marks}} & \text{is} & \underbrace{\quad \text{160}\quad} \\ a + b & = & 160 \end{array}$

The first fact is equivalent to this equation: $a + b = 160$ .

STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

Similarly, we can identify key parts of the second fact, and translate each into an expression.

\begin{matrix} \underset{⏟}{Anthea's mark} & is & \underset{⏟}{8 less than Bokamoso's mark} \\ a & = & b - 8 \end{matrix}

$\begin{array}{c} \underbrace{\text{Anthea's mark}} & \text{is} & \underbrace{\text{8 less than Bokamoso's mark}} \\ a & = & b - 8 \end{array}$

This equation, $a = b - 8$ , means that Anthea's mark is less that Bokamoso's mark.

The correct answers are:

Fact about the test scores	Equation
The sum of the marks is $\text{160}$ .	$a + b = 160$
Anthea's mark is $\text{8}$ less than Bokamoso's mark.	$a = b - 8$

Submit your answer as: and

Exercises

Setting up simultaneous equations

Here are facts about two numbers:

The sum of the numbers is $\text{15}$ .
The numbers are consecutive numbers.

Answer:

Fact about the numbers	Equation
The sum of the numbers is $\text{15}$ .
The numbers are consecutive numbers.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.

STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

Word	Meaning
sum	+
product	$\times$
is	=
consecutive	1 apart
more than	add to
less than	subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let $n_{1}$ represent one of the numbers and $n_{2}$ represent the other number". So we can break up the first fact like this:

\begin{array}{ccc} \underset{⏟}{The sum of the numbers} & is & \underset{⏟}{15} \\ n_{1} + n_{2} & = & 15 \end{array}

$\begin{array}{ccc} \underbrace{\text{The sum of the numbers}} & \text{is} & \underbrace{\quad\text{15}\quad} \\ n_{1} + n_{2} & = & 15 \end{array}$

The correct equation for the first fact is $n_{1} + n_{2} = 15$ .

STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

\begin{array}{ccc} \underset{⏟}{The larger number} & is & \underset{⏟}{1 more than the smaller number} \\ n_{2} & = & n_{1} + 1 \end{array}

$\begin{array}{ccc} \underbrace{\text{The larger number}} & \text{is} & \underbrace{\text{1 more than the smaller number}} \\ n_{2} & = & n_{1} + 1 \end{array}$

This makes sense: consecutive numbers are separated by $\text{1}$ . The equation shows that with the $\text{+1}$ .

The correct answers are:

Fact about the numbers	Equation
The sum of the numbers is $\text{15}$ .	$n_1 + n_2 = 15$
The numbers are consecutive numbers.	$n_{2} = n_{1} + 1$

Submit your answer as: and

Setting up simultaneous equations

Here are facts about two numbers:

The product of the numbers is $\text{72}$ .
The numbers are consecutive numbers.

Answer:

Fact about the numbers	Equation
The product of the numbers is $\text{72}$ .
The numbers are consecutive numbers.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is product. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.

STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

Word	Meaning
sum	+
product	$\times$
is	=
consecutive	1 apart
more than	add to
less than	subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let $n_{1}$ represent one of the numbers and $n_{2}$ represent the other number". So we can break up the first fact like this:

\begin{array}{ccc} \underset{⏟}{The product of the numbers} & is & \underset{⏟}{72} \\ n_{1} \times n_{2} & = & 72 \end{array}

$\begin{array}{ccc} \underbrace{\text{The product of the numbers}} & \text{is} & \underbrace{\quad\text{72}\quad} \\ n_{1} \times n_{2} & = & 72 \end{array}$

The correct equation for the first fact is $n_{1} \times n_{2} = 72$ .

STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

\begin{array}{ccc} \underset{⏟}{The smaller number} & is & \underset{⏟}{1 less than the larger number} \\ n_{1} & = & n_{2} - 1 \end{array}

$\begin{array}{ccc} \underbrace{\text{The smaller number}} & \text{is} & \underbrace{\text{1 less than the larger number}} \\ n_{1} & = & n_{2} - 1 \end{array}$

This makes sense: consecutive numbers are separated by $\text{1}$ . The equation shows that with the $-\text{1}$ .

The correct answers are:

Fact about the numbers	Equation
The product of the numbers is $\text{72}$ .	$n_1 \times n_2 = 72$
The numbers are consecutive numbers.	$n_{1} = n_{2} - 1$

Submit your answer as: and

Setting up simultaneous equations

Here are facts about two numbers:

The product of the numbers is $\text{30}$ .
The numbers are consecutive numbers.

Answer:

Fact about the numbers	Equation
The product of the numbers is $\text{30}$ .
The numbers are consecutive numbers.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is product. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.

STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

Word	Meaning
sum	+
product	$\times$
is	=
consecutive	1 apart
more than	add to
less than	subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let $n_{1}$ represent one of the numbers and $n_{2}$ represent the other number". So we can break up the first fact like this:

\begin{array}{ccc} \underset{⏟}{The product of the numbers} & is & \underset{⏟}{30} \\ n_{1} \times n_{2} & = & 30 \end{array}

$\begin{array}{ccc} \underbrace{\text{The product of the numbers}} & \text{is} & \underbrace{\quad\text{30}\quad} \\ n_{1} \times n_{2} & = & 30 \end{array}$

The correct equation for the first fact is $n_{1} \times n_{2} = 30$ .

STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

\begin{array}{ccc} \underset{⏟}{The larger number} & is & \underset{⏟}{1 more than the smaller number} \\ n_{2} & = & n_{1} + 1 \end{array}

$\begin{array}{ccc} \underbrace{\text{The larger number}} & \text{is} & \underbrace{\text{1 more than the smaller number}} \\ n_{2} & = & n_{1} + 1 \end{array}$

This makes sense: consecutive numbers are separated by $\text{1}$ . The equation shows that with the $\text{+1}$ .

The correct answers are:

Fact about the numbers	Equation
The product of the numbers is $\text{30}$ .	$n_1 \times n_2 = 30$
The numbers are consecutive numbers.	$n_{2} = n_{1} + 1$

Submit your answer as: and

Word problems: distance, speed and time

Two missiles start out at two different launchers. The missiles are $\text{1 232}\ \text{km}$ apart. The missiles start moving towards each other. One missile is moving at $\text{291}\ \text{km/h}$ and the other missile at $\text{325}\ \text{km/h}$ .

If both missiles started their journey at the same time, how long will they take to pass each other?

Answer: The two missiles pass each other after hours.

numeric

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.

STEP: Sketch a simple picture to organise the information
[−2 points ⇒ 4 / 6 points left]

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two missiles must be equal to the total distance when the missiles meet:

\begin{aligned} d_{1} + d_{2} & = d_{t o t a l} \\ ↓ \\ d_{1} + d_{2} & = 1 232 km \end{aligned}

$\begin{aligned} d_{1} + d_{2} & = d_{total} \\ \quad & \downarrow \quad \\ d_{1} + d_{2} & = \text{1 232}\ \text{km} \end{aligned}$

STEP: Write equations to describe the motion of the missiles
[−2 points ⇒ 2 / 6 points left]

This question is about distances, speeds, and times. The equation connecting these values is

\begin{aligned} speed & = \frac{distance}{time} \\ OR \\ distance & = speed \times time \end{aligned}

$\begin{aligned} \text{speed } & = \frac{\text{distance }}{\text{time}} \\ \quad & \text{ OR } \quad \\ \text{distance } & = \text{speed } \times \text{time} \end{aligned}$

We want to know the amount of time needed for the missiles to meet. Let the time taken be $t$ . Then we can write an expression for the distance each of the missiles travels:

For missile 1:

\begin{aligned} d_{1} & = s_{1} t \\ _{missile is 291 km/h}^{The speed of the first} \to d_{1} & = 291 t \end{aligned}

$\begin{aligned} d_{1} &= s_{1} t \\ ^{\text{The speed of the first}}_{\text{missile is } \text{291}\ \text{km/h}} \quad \rightarrow \quad d_{1} &= \text{291}t \end{aligned}$

For missile 2:

\begin{aligned} d_{2} & = s_{2} t \\ _{missile is 325 km/h}^{The speed of the second} \to d_{2} & = 325 t \end{aligned}

$\begin{aligned} d_{2} &= s_{2} t \\ ^{\text{The speed of the second}}_{\text{missile is } \text{325}\ \text{km/h}} \quad \rightarrow \quad d_{2} &= \text{325}t \end{aligned}$

STEP: Solve the equations simultaneously for

t

$t$
[−2 points ⇒ 0 / 6 points left]

Now we have three different equations. We can combine them using substitution to solve for the value of $t$ .

\begin{aligned} d_{1} + d_{2} & = 1 232 \\ (291 t) + (325 t) & = 1 232 \\ 616 t & = 1 232 \\ t & = \frac{1 232}{616} \\ t & = 2 \end{aligned}

$\begin{aligned} d_{1} + d_{2} &= \text{1 232} \\ (\text{291}t) + (\text{325}t) &= \text{1 232} \\ \text{616}t &= \text{1 232} \\ t &= \frac{\text{1 232}}{\text{616}} \\ t &= \text{2} \end{aligned}$

The missiles will meet after $\text{2}$ hours.

Submit your answer as:

Word problems: distance, speed and time

Two planes start out at two different airports. The planes are $\text{2 044}\ \text{km}$ apart. The planes start flying towards each other. One plane is flying at $\text{311}\ \text{km/h}$ and the other plane at $\text{200}\ \text{km/h}$ .

If both planes started their journey at the same time, how long will they take to pass each other?

Answer: The two planes pass each other after hours.

numeric

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.

STEP: Sketch a simple picture to organise the information
[−2 points ⇒ 4 / 6 points left]

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two planes must be equal to the total distance when the planes meet:

\begin{aligned} d_{1} + d_{2} & = d_{t o t a l} \\ ↓ \\ d_{1} + d_{2} & = 2 044 km \end{aligned}

$\begin{aligned} d_{1} + d_{2} & = d_{total} \\ \quad & \downarrow \quad \\ d_{1} + d_{2} & = \text{2 044}\ \text{km} \end{aligned}$

STEP: Write equations to describe the motion of the planes
[−2 points ⇒ 2 / 6 points left]

This question is about distances, speeds, and times. The equation connecting these values is

\begin{aligned} speed & = \frac{distance}{time} \\ OR \\ distance & = speed \times time \end{aligned}

$\begin{aligned} \text{speed } & = \frac{\text{distance }}{\text{time}} \\ \quad & \text{ OR } \quad \\ \text{distance } & = \text{speed } \times \text{time} \end{aligned}$

We want to know the amount of time needed for the planes to meet. Let the time taken be $t$ . Then we can write an expression for the distance each of the planes travels:

For plane 1:

\begin{aligned} d_{1} & = s_{1} t \\ _{plane is 311 km/h}^{The speed of the first} \to d_{1} & = 311 t \end{aligned}

$\begin{aligned} d_{1} &= s_{1} t \\ ^{\text{The speed of the first}}_{\text{plane is } \text{311}\ \text{km/h}} \quad \rightarrow \quad d_{1} &= \text{311}t \end{aligned}$

For plane 2:

\begin{aligned} d_{2} & = s_{2} t \\ _{plane is 200 km/h}^{The speed of the second} \to d_{2} & = 200 t \end{aligned}

$\begin{aligned} d_{2} &= s_{2} t \\ ^{\text{The speed of the second}}_{\text{plane is } \text{200}\ \text{km/h}} \quad \rightarrow \quad d_{2} &= \text{200}t \end{aligned}$

STEP: Solve the equations simultaneously for

t

$t$
[−2 points ⇒ 0 / 6 points left]

Now we have three different equations. We can combine them using substitution to solve for the value of $t$ .

\begin{aligned} d_{1} + d_{2} & = 2 044 \\ (311 t) + (200 t) & = 2 044 \\ 511 t & = 2 044 \\ t & = \frac{2 044}{511} \\ t & = 4 \end{aligned}

$\begin{aligned} d_{1} + d_{2} &= \text{2 044} \\ (\text{311}t) + (\text{200}t) &= \text{2 044} \\ \text{511}t &= \text{2 044} \\ t &= \frac{\text{2 044}}{\text{511}} \\ t &= \text{4} \end{aligned}$

The planes will meet after $\text{4}$ hours.

Submit your answer as:

Word problems: distance, speed and time

Two trains start out at two different stations. The trains are $\text{3 195}\ \text{km}$ apart. The trains start travelling towards each other. One train is travelling at $\text{337}\ \text{km/h}$ and the other train at $\text{302}\ \text{km/h}$ .

If both trains started their journey at the same time, how long will they take to pass each other?

Answer: The two trains pass each other after hours.

numeric

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.

STEP: Sketch a simple picture to organise the information
[−2 points ⇒ 4 / 6 points left]

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two trains must be equal to the total distance when the trains meet:

\begin{aligned} d_{1} + d_{2} & = d_{t o t a l} \\ ↓ \\ d_{1} + d_{2} & = 3 195 km \end{aligned}

$\begin{aligned} d_{1} + d_{2} & = d_{total} \\ \quad & \downarrow \quad \\ d_{1} + d_{2} & = \text{3 195}\ \text{km} \end{aligned}$

STEP: Write equations to describe the motion of the trains
[−2 points ⇒ 2 / 6 points left]

This question is about distances, speeds, and times. The equation connecting these values is

\begin{aligned} speed & = \frac{distance}{time} \\ OR \\ distance & = speed \times time \end{aligned}

$\begin{aligned} \text{speed } & = \frac{\text{distance }}{\text{time}} \\ \quad & \text{ OR } \quad \\ \text{distance } & = \text{speed } \times \text{time} \end{aligned}$

We want to know the amount of time needed for the trains to meet. Let the time taken be $t$ . Then we can write an expression for the distance each of the trains travels:

For train 1:

\begin{aligned} d_{1} & = s_{1} t \\ _{train is 337 km/h}^{The speed of the first} \to d_{1} & = 337 t \end{aligned}

$\begin{aligned} d_{1} &= s_{1} t \\ ^{\text{The speed of the first}}_{\text{train is } \text{337}\ \text{km/h}} \quad \rightarrow \quad d_{1} &= \text{337}t \end{aligned}$

For train 2:

\begin{aligned} d_{2} & = s_{2} t \\ _{train is 302 km/h}^{The speed of the second} \to d_{2} & = 302 t \end{aligned}

$\begin{aligned} d_{2} &= s_{2} t \\ ^{\text{The speed of the second}}_{\text{train is } \text{302}\ \text{km/h}} \quad \rightarrow \quad d_{2} &= \text{302}t \end{aligned}$

STEP: Solve the equations simultaneously for

t

$t$
[−2 points ⇒ 0 / 6 points left]

Now we have three different equations. We can combine them using substitution to solve for the value of $t$ .

\begin{aligned} d_{1} + d_{2} & = 3 195 \\ (337 t) + (302 t) & = 3 195 \\ 639 t & = 3 195 \\ t & = \frac{3 195}{639} \\ t & = 5 \end{aligned}

$\begin{aligned} d_{1} + d_{2} &= \text{3 195} \\ (\text{337}t) + (\text{302}t) &= \text{3 195} \\ \text{639}t &= \text{3 195} \\ t &= \frac{\text{3 195}}{\text{639}} \\ t &= \text{5} \end{aligned}$

The trains will meet after $\text{5}$ hours.

Submit your answer as:

Word problems: solving simultaneous equations

A group of friends is buying lunch together. The group buys $\text{4}$ waffles and $\text{5}$ sandwiches. Here are some facts about their lunch:

the total cost for the $\text{4}$ waffles and $\text{5}$ sandwiches is $\text{R}\,\text{276}$
a waffle costs $\text{R}\,\text{6}$ more than a sandwich

What is the price for one waffle and the price for one sandwich?

Answer:

A waffle costs $\text{R}$ and a sandwich costs $\text{R}$ .

numeric

HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

You need to choose variables to represent the things you want to find and write equations based on the information in the question.

STEP: Pick variables for the things we want to know
[−1 point ⇒ 6 / 7 points left]

The two things we want to know in this question are the prices for each of the items (a waffle and a sandwich). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

\begin{aligned} w & = the price of a waffle \\ s & = the price of a sandwich \end{aligned}

$\begin{aligned} w & = \text{the price of a waffle} \\ s & = \text{the price of a sandwich} \end{aligned}$

STEP: Write equations based on the information in the question
[−2 points ⇒ 4 / 7 points left]

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the $\text{4}$ waffles and $\text{5}$ sandwiches is $\text{R}\,\text{276}$ ." We can use the expression $4 w$ to represent the price of the $\text{4}$ waffles. Similarly, the expression $5 s$ represents the price of the $\text{5}$ sandwiches. With these values we can write a full equation for the prices:

\begin{aligned} In words: & the total cost for the 4 waffles and 5 sandwiches is R 276 \\ In maths: & 4 w + 5 s = 276 \end{aligned}

$\begin{aligned} \text{In words: } & \;\; \text{the total cost for the 4 waffles and 5 sandwiches is } \text{R}\,\text{276} \\ \text{In maths: } & \;\; 4 w + 5 s = 276 \end{aligned}$

Now we can use the second point. It says, "a waffle costs $\text{R}\,\text{6}$ more than a sandwich." We can write this as an equation like this:

\begin{aligned} In words: & a waffle costs R 6 more than a sandwich \\ In maths: & w = s + 6 \end{aligned}

$\begin{aligned} \text{In words: } & \;\; \text{a waffle costs } \text{R}\,\text{6} \text{ more than a sandwich} \\ \text{In maths: } & \;\; w = s+6 \end{aligned}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 2 / 7 points left]

Now that we have two equations, we need to solve them simultaneously.

\begin{aligned} 4 w + 5 s & = 276 \\ w & = s + 6 \end{aligned}

$\begin{aligned} 4 w + 5 s & = 276 \\ w & = s+6 \end{aligned}$

We could use elimination, but substitution is a better choice (because $w$ is already the subject). Substitute the second equation into the first equation and solve!

\begin{aligned} 4 w + 5 s & = 276 \\ 4 (s + 6) + 5 s & = 276 \\ 4 s + 24 + 5 s & = 276 \\ 9 s & = 276 - 24 \\ s & = \frac{252}{9} \\ = 28 \end{aligned}

$\begin{aligned} 4 w + 5 s &= 276 \\ 4 \left( s+6 \right) +5 s &= 276 \\ 4 s + 24 + 5 s &= 276 \\ 9 s &= 276 - 24 \\ s &= \frac{252} {9} \\ &= 28 \end{aligned}$

This means that the price of one sandwich is $\text{R}\,\text{28}$ .

STEP: Find the other variable's value
[−1 point ⇒ 1 / 7 points left]

Finally, use the value we found for $s$ to find the value of $w$ (the price of a waffle). We can use either equation to do this, but the second one is easier to use (because $w$ is already isolated).

\begin{aligned} w & = s + 6 \\ = 28 + 6 \\ = 34 \end{aligned}

$\begin{aligned} w &= s+6 \\ &= 28+6 \\ &= 34 \end{aligned}$

The price for one waffle is $\text{R}\,\text{34}$ .

STEP: Write the final answer
[−1 point ⇒ 0 / 7 points left]

It is important to write the answer to a word problem as a complete sentence.

The price of the waffle is $\text{R}\,\text{34}$ while a sandwich costs $\text{R}\,\text{28}$ .

Submit your answer as: and

Word problems: solving simultaneous equations

A group of friends is buying lunch together. The group buys $\text{4}$ sandwiches and $\text{2}$ pizzas. Here are some facts about their lunch:

the total cost for the $\text{4}$ sandwiches and $\text{2}$ pizzas is $\text{R}\,\text{198}$
a sandwich costs $\text{R}\,\text{6}$ more than a pizza

What is the price for one sandwich and the price for one pizza?

Answer:

A sandwich costs $\text{R}$ and a pizza costs $\text{R}$ .

numeric

HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

You need to choose variables to represent the things you want to find and write equations based on the information in the question.

STEP: Pick variables for the things we want to know
[−1 point ⇒ 6 / 7 points left]

The two things we want to know in this question are the prices for each of the items (a sandwich and a pizza). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

\begin{aligned} s & = the price of a sandwich \\ p & = the price of a pizza \end{aligned}

$\begin{aligned} s & = \text{the price of a sandwich} \\ p & = \text{the price of a pizza} \end{aligned}$

STEP: Write equations based on the information in the question
[−2 points ⇒ 4 / 7 points left]

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the $\text{4}$ sandwiches and $\text{2}$ pizzas is $\text{R}\,\text{198}$ ." We can use the expression $4 s$ to represent the price of the $\text{4}$ sandwiches. Similarly, the expression $2 p$ represents the price of the $\text{2}$ pizzas. With these values we can write a full equation for the prices:

\begin{aligned} In words: & the total cost for the 4 sandwiches and 2 pizzas is R 198 \\ In maths: & 4 s + 2 p = 198 \end{aligned}

$\begin{aligned} \text{In words: } & \;\; \text{the total cost for the 4 sandwiches and 2 pizzas is } \text{R}\,\text{198} \\ \text{In maths: } & \;\; 4 s + 2 p = 198 \end{aligned}$

Now we can use the second point. It says, "a sandwich costs $\text{R}\,\text{6}$ more than a pizza." We can write this as an equation like this:

\begin{aligned} In words: & a sandwich costs R 6 more than a pizza \\ In maths: & s = p + 6 \end{aligned}

$\begin{aligned} \text{In words: } & \;\; \text{a sandwich costs } \text{R}\,\text{6} \text{ more than a pizza} \\ \text{In maths: } & \;\; s = p+6 \end{aligned}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 2 / 7 points left]

Now that we have two equations, we need to solve them simultaneously.

\begin{aligned} 4 s + 2 p & = 198 \\ s & = p + 6 \end{aligned}

$\begin{aligned} 4 s + 2 p & = 198 \\ s & = p+6 \end{aligned}$

We could use elimination, but substitution is a better choice (because $s$ is already the subject). Substitute the second equation into the first equation and solve!

\begin{aligned} 4 s + 2 p & = 198 \\ 4 (p + 6) + 2 p & = 198 \\ 4 p + 24 + 2 p & = 198 \\ 6 p & = 198 - 24 \\ p & = \frac{174}{6} \\ = 29 \end{aligned}

$\begin{aligned} 4 s + 2 p &= 198 \\ 4 \left( p+6 \right) +2 p &= 198 \\ 4 p + 24 + 2 p &= 198 \\ 6 p &= 198 - 24 \\ p &= \frac{174} {6} \\ &= 29 \end{aligned}$

This means that the price of one pizza is $\text{R}\,\text{29}$ .

STEP: Find the other variable's value
[−1 point ⇒ 1 / 7 points left]

Finally, use the value we found for $p$ to find the value of $s$ (the price of a sandwich). We can use either equation to do this, but the second one is easier to use (because $s$ is already isolated).

\begin{aligned} s & = p + 6 \\ = 29 + 6 \\ = 35 \end{aligned}

$\begin{aligned} s &= p+6 \\ &= 29+6 \\ &= 35 \end{aligned}$

The price for one sandwich is $\text{R}\,\text{35}$ .

STEP: Write the final answer
[−1 point ⇒ 0 / 7 points left]

It is important to write the answer to a word problem as a complete sentence.

The price of the sandwich is $\text{R}\,\text{35}$ while a pizza costs $\text{R}\,\text{29}$ .

Submit your answer as: and

Word problems: solving simultaneous equations

A group of friends is buying lunch together. The group buys $\text{6}$ milkshakes and $\text{4}$ waffles. Here are some facts about their lunch:

the total cost for the $\text{6}$ milkshakes and $\text{4}$ waffles is $\text{R}\,\text{264}$
a milkshake costs $\text{R}\,\text{4}$ more than a waffle

What is the price for one milkshake and the price for one waffle?

Answer:

A milkshake costs $\text{R}$ and a waffle costs $\text{R}$ .

numeric

HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

You need to choose variables to represent the things you want to find and write equations based on the information in the question.

STEP: Pick variables for the things we want to know
[−1 point ⇒ 6 / 7 points left]

The two things we want to know in this question are the prices for each of the items (a milkshake and a waffle). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

\begin{aligned} m & = the price of a milkshake \\ w & = the price of a waffle \end{aligned}

$\begin{aligned} m & = \text{the price of a milkshake} \\ w & = \text{the price of a waffle} \end{aligned}$

STEP: Write equations based on the information in the question
[−2 points ⇒ 4 / 7 points left]

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the $\text{6}$ milkshakes and $\text{4}$ waffles is $\text{R}\,\text{264}$ ." We can use the expression $6 m$ to represent the price of the $\text{6}$ milkshakes. Similarly, the expression $4 w$ represents the price of the $\text{4}$ waffles. With these values we can write a full equation for the prices:

\begin{aligned} In words: & the total cost for the 6 milkshakes and 4 waffles is R 264 \\ In maths: & 6 m + 4 w = 264 \end{aligned}

$\begin{aligned} \text{In words: } & \;\; \text{the total cost for the 6 milkshakes and 4 waffles is } \text{R}\,\text{264} \\ \text{In maths: } & \;\; 6 m + 4 w = 264 \end{aligned}$

Now we can use the second point. It says, "a milkshake costs $\text{R}\,\text{4}$ more than a waffle." We can write this as an equation like this:

\begin{aligned} In words: & a milkshake costs R 4 more than a waffle \\ In maths: & m = w + 4 \end{aligned}

$\begin{aligned} \text{In words: } & \;\; \text{a milkshake costs } \text{R}\,\text{4} \text{ more than a waffle} \\ \text{In maths: } & \;\; m = w+4 \end{aligned}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 2 / 7 points left]

Now that we have two equations, we need to solve them simultaneously.

\begin{aligned} 6 m + 4 w & = 264 \\ m & = w + 4 \end{aligned}

$\begin{aligned} 6 m + 4 w & = 264 \\ m & = w+4 \end{aligned}$

We could use elimination, but substitution is a better choice (because $m$ is already the subject). Substitute the second equation into the first equation and solve!

\begin{aligned} 6 m + 4 w & = 264 \\ 6 (w + 4) + 4 w & = 264 \\ 6 w + 24 + 4 w & = 264 \\ 10 w & = 264 - 24 \\ w & = \frac{240}{10} \\ = 24 \end{aligned}

$\begin{aligned} 6 m + 4 w &= 264 \\ 6 \left( w+4 \right) +4 w &= 264 \\ 6 w + 24 + 4 w &= 264 \\ 10 w &= 264 - 24 \\ w &= \frac{240} {10} \\ &= 24 \end{aligned}$

This means that the price of one waffle is $\text{R}\,\text{24}$ .

STEP: Find the other variable's value
[−1 point ⇒ 1 / 7 points left]

Finally, use the value we found for $w$ to find the value of $m$ (the price of a milkshake). We can use either equation to do this, but the second one is easier to use (because $m$ is already isolated).

\begin{aligned} m & = w + 4 \\ = 24 + 4 \\ = 28 \end{aligned}

$\begin{aligned} m &= w+4 \\ &= 24+4 \\ &= 28 \end{aligned}$

The price for one milkshake is $\text{R}\,\text{28}$ .

STEP: Write the final answer
[−1 point ⇒ 0 / 7 points left]

It is important to write the answer to a word problem as a complete sentence.

The price of the milkshake is $\text{R}\,\text{28}$ while a waffle costs $\text{R}\,\text{24}$ .

Submit your answer as: and

Word problems: checking answers about shopping

Suppose you must solve this word problem:

At a shop there are some fruits and sweets for sale. Ayodeji buys some bananas and some chocolates. The total cost for $\text{5}$ bananas and $\text{5}$ chocolates is $\text{R}\,\text{48,00}$ . And each chocolate costs $\text{R}\,\text{0,60}$ less than each banana. What is the price for one banana?

The answer to this question is a number. Answer the two questions below about this number.

Answer:

Can the price be a decimal number (or must it be an integer)?
Can the price be a negative number?

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: Decide if the price can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

The answer for the first question is: Yes, the price can be a decimal number.

STEP: Decide if the price can be negative
[−1 point ⇒ 0 / 2 points left]

The answer to the second question is: No, the price cannot be a negative number.

The correct answer choices are:

Yes, it can be a decimal
No, it cannot be negative

Submit your answer as: and

Word problems: checking answers about shopping

Suppose you must solve this word problem:

At a shop there are some fruits and sweets for sale. Yaseen buys some mangoes and some packs of gum. The total cost for $\text{2}$ mangoes and $\text{4}$ packs of gum is $\text{R}\,\text{34,80}$ . And each pack of gum costs $\text{R}\,\text{0,30}$ less than each mango. What is the price for one mango?

The answer to this question is a number. Answer the two questions below about this number.

Answer:

Can the price be a decimal number (or must it be an integer)?
Can the price be a negative number?

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the price of each mango. Can the price be a decimal, like $\text{4,5}$ ? Can it be negative, like $-\text{5}$ ?

STEP: Decide if the price can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

The answer for the first question is: Yes, the price can be a decimal number.

STEP: Decide if the price can be negative
[−1 point ⇒ 0 / 2 points left]

The answer to the second question is: No, the price cannot be a negative number.

The correct answer choices are:

Yes, it can be a decimal
No, it cannot be negative

Submit your answer as: and

Word problems: checking answers about shopping

Suppose you must solve this word problem:

At a shop there are some fruits and sweets for sale. Ismail buys some bananas and some packs of gum. The total cost for $\text{2}$ bananas and $\text{5}$ packs of gum is $\text{R}\,\text{29,10}$ . And each pack of gum costs $\text{R}\,\text{2,30}$ less than each banana. What is the price for one banana?

The answer to this question is a number. Answer the two questions below about this number.

Answer:

Can the price be a decimal number (or must it be an integer)?
Can the price be a negative number?

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

STEP: Decide if the price can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

The answer for the first question is: Yes, the price can be a decimal number.

STEP: Decide if the price can be negative
[−1 point ⇒ 0 / 2 points left]

The answer to the second question is: No, the price cannot be a negative number.

The correct answer choices are:

Yes, it can be a decimal
No, it cannot be negative

Submit your answer as: and

Word problems: checking answers with shapes

Suppose you must solve this word problem:

Ntombebhongo is drawing a pattern of rectangles in her notebook. The number of rectangles in each figure is $\text{3}$ more than the number of rectangles in the previous figure. If there are $\text{12}$ rectangles in the third figure, how many rectangles are there in the first figure?

The answer to this question is a number. Answer the two questions below about this number.

Answer:

Can the number of rectangles be a decimal number (or must it be an integer)?
Can the number of rectangles be a negative number?

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the number of rectangles in the first figure of Ntombebhongo's diagram. Can the number of rectangles be a decimal, like $\text{4,5}$ ? Can it be negative, like $-\text{5}$ ?

STEP: Decide if the number of rectangles can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

For the first question we must decide if the number of rectangles can be a decimal number. Is that fine or is it impossible?

The answer for the first question is: No, it must be an integer, the number of rectangles cannot be a decimal number.

STEP: Decide if the number of rectangles can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if number of rectangles can be negative, or must be positive. The number of rectangles cannot be negative. If Ntombebhongo drew any rectangles at all, then the answer to this question must be positive. There is no such thing as $-\text{4}$ rectangles.

The answer for the second question is: No, the number of rectangles cannot be a negative number.

The correct answer choices are:

No, it must be an integer
No, it cannot be negative

Submit your answer as: and

Word problems: checking answers with shapes

Suppose you must solve this word problem:

Chris is drawing a rectangle in his notebook. The length of the rectangle is $\text{5}\ \text{cm}$ more than its width. The area of the rectangle is $\text{6}\ \text{cm$^{2}$}$ . What is the width of the rectangle?

The answer to this question is a number. Answer the two questions below about this number.

Answer:

Can the width of the rectangle be a decimal number (or must it be an integer)?
Can the width of the rectangle be a negative number?

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the width of the rectangle. Can the width of the rectangle be a decimal, like $\text{4,5}$ ? Can it be negative, like $-\text{5}$ ?

STEP: Decide if the width of the rectangle can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

For the first question we must decide if the width of the rectangle can be a decimal number. Is that fine or is it impossible?

In other words, can a distance be a decimal number like $\text{3,5}\ \text{cm}$ , or must a distance be a whole number like $\text{4}\ \text{cm}$ ? The answer is that there is no reason a distance must be limited to whole number values. This is different from the number of days in the school year. It is not possible to have $\text{3,5}$ days of school, but a rectangle certainly can have a width of $\text{3,5}\ \text{cm}$ . Or $\text{5,927}\ \text{cm}$ . There is no reason why the width of the rectangle must be a whole number.

The answer for the first question is: Yes, it can be a decimal, the width of the rectangle can be a decimal number.

STEP: Decide if the width of the rectangle can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if width of the rectangle can be negative, or must be positive. This is straightfoward: distances cannot be negative. Imagine a room $-\text{5}$ metres wide. No way!

The answer for the second question is: No, the width of the rectangle cannot be a negative number.

The correct answer choices are:

Yes, it can be a decimal
No, it cannot be negative

Submit your answer as: and

Word problems: checking answers with shapes

Suppose you must solve this word problem:

Habubakar is drawing a pattern of rectangles in his notebook. The number of rectangles in each figure is $\text{2}$ more than the number of rectangles in the previous figure. If there are $\text{11}$ rectangles in the fourth figure, how many rectangles are there in the first figure?

The answer to this question is a number. Answer the two questions below about this number.

Answer:

Can the number of rectangles be a decimal number (or must it be an integer)?
Can the number of rectangles be a negative number?

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You need to think about the meaning of the number you want. The question asks us for the number of rectangles in the first figure of Habubakar's diagram. Can the number of rectangles be a decimal, like $\text{4,5}$ ? Can it be negative, like $-\text{5}$ ?

STEP: Decide if the number of rectangles can be a decimal (non-integer) value
[−1 point ⇒ 1 / 2 points left]

For the first question we must decide if the number of rectangles can be a decimal number. Is that fine or is it impossible?

The answer for the first question is: No, it must be an integer, the number of rectangles cannot be a decimal number.

STEP: Decide if the number of rectangles can be negative
[−1 point ⇒ 0 / 2 points left]

The second question is similar, but this time we need to decide if number of rectangles can be negative, or must be positive. The number of rectangles cannot be negative. If Habubakar drew any rectangles at all, then the answer to this question must be positive. There is no such thing as $-\text{4}$ rectangles.

The answer for the second question is: No, the number of rectangles cannot be a negative number.

The correct answer choices are:

No, it must be an integer
No, it cannot be negative

Submit your answer as: and

Word problems: test scores and simultaneous equations

Louis and Musa are friends. Louis takes Musa's physics test paper and says: “I have $\text{14}$ marks more than you do and the sum of both our marks is equal to $\text{166}$ . What are our marks?”

Answer:

Louis got marks and Musa got marks for the physics test paper.

numeric

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

STEP: Pick variables for the marks
[−1 point ⇒ 5 / 6 points left]

\begin{aligned} l & = Louis's mark \\ m & = Musa's mark \end{aligned}

$\begin{aligned} l &= \text{ Louis's mark} \\ m &= \text{ Musa's mark} \end{aligned}$

STEP: Write equations about the students' marks
[−2 points ⇒ 3 / 6 points left]

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

\begin{aligned} _{more than Musa}^{Louis has 14 marks} & \to l = m + 14 \\ _{(the total) is 166}^{The sum of the marks} & \to l + m = 166 \end{aligned}

$\begin{aligned} ^{\text{Louis has } \text{14} \text{ marks}} _{\text{ more than Musa}} \quad & \rightarrow \quad l = m + \text{14} \\ \\ ^{\text{The sum of the marks}} _{\text{(the total) is } \text{166}} \quad & \rightarrow \quad l+m = \text{166} \end{aligned}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

\begin{aligned} l + m & = 166 \\ (m + 14) + m & = 166 \\ 2 m & = 166 - 14 \\ m & = \frac{152}{2} \\ = 76 \end{aligned}

$\begin{aligned} l+m &= \text{166} \\ (m+\text{14})+m &= \text{166} \\ 2m &= 166 - \text{14}\\ m &= \frac{\text{152}}{\text{2}}\\ &= \text{76} \end{aligned}$

This means that Musa's mark is $\text{76}$ .

STEP: Find Louis's mark
[−1 point ⇒ 0 / 6 points left]

Now we can use Musa's mark and one of the equations we have to find Louis's mark. The easiest way to do that is to substitute Musa's mark back into the first equation:

\begin{aligned} l & = m + 14 \\ = (76) + 14 \\ = 90 \end{aligned}

$\begin{aligned} l &= m+\text{14} \\ &= \left( \text{76} \right) + \text{14} \\ &= \text{90} \end{aligned}$

The students achieved these marks: Louis earned $\text{90}$ and Musa earned $\text{76}$ .

Submit your answer as: and

Word problems: test scores and simultaneous equations

Nicole and Gabisile are friends. Nicole takes Gabisile's chemistry test paper and says: “I have $\text{19}$ marks more than you do and the sum of both our marks is equal to $\text{117}$ . What are our marks?”

Answer:

Nicole got marks and Gabisile got marks for the chemistry test paper.

numeric

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

STEP: Pick variables for the marks
[−1 point ⇒ 5 / 6 points left]

\begin{aligned} n & = Nicole's mark \\ g & = Gabisile's mark \end{aligned}

$\begin{aligned} n &= \text{ Nicole's mark} \\ g &= \text{ Gabisile's mark} \end{aligned}$

STEP: Write equations about the students' marks
[−2 points ⇒ 3 / 6 points left]

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

\begin{aligned} _{more than Gabisile}^{Nicole has 19 marks} & \to n = g + 19 \\ _{(the total) is 117}^{The sum of the marks} & \to n + g = 117 \end{aligned}

$\begin{aligned} ^{\text{Nicole has } \text{19} \text{ marks}} _{\text{ more than Gabisile}} \quad & \rightarrow \quad n = g + \text{19} \\ \\ ^{\text{The sum of the marks}} _{\text{(the total) is } \text{117}} \quad & \rightarrow \quad n+g = \text{117} \end{aligned}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

\begin{aligned} n + g & = 117 \\ (g + 19) + g & = 117 \\ 2 g & = 117 - 19 \\ g & = \frac{98}{2} \\ = 49 \end{aligned}

$\begin{aligned} n+g &= \text{117} \\ (g+\text{19})+g &= \text{117} \\ 2g &= 117 - \text{19}\\ g &= \frac{\text{98}}{\text{2}}\\ &= \text{49} \end{aligned}$

This means that Gabisile's mark is $\text{49}$ .

STEP: Find Nicole's mark
[−1 point ⇒ 0 / 6 points left]

Now we can use Gabisile's mark and one of the equations we have to find Nicole's mark. The easiest way to do that is to substitute Gabisile's mark back into the first equation:

\begin{aligned} n & = g + 19 \\ = (49) + 19 \\ = 68 \end{aligned}

$\begin{aligned} n &= g+\text{19} \\ &= \left( \text{49} \right) + \text{19} \\ &= \text{68} \end{aligned}$

The students achieved these marks: Nicole earned $\text{68}$ and Gabisile earned $\text{49}$ .

Submit your answer as: and

Word problems: test scores and simultaneous equations

Bianca and Carl are friends. Bianca takes Carl's mathematics test paper and says: “I have $\text{19}$ marks more than you do and the sum of both our marks is equal to $\text{153}$ . What are our marks?”

Answer:

Bianca got marks and Carl got marks for the mathematics test paper.

numeric

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

STEP: Pick variables for the marks
[−1 point ⇒ 5 / 6 points left]

\begin{aligned} b & = Bianca's mark \\ c & = Carl's mark \end{aligned}

$\begin{aligned} b &= \text{ Bianca's mark} \\ c &= \text{ Carl's mark} \end{aligned}$

STEP: Write equations about the students' marks
[−2 points ⇒ 3 / 6 points left]

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

\begin{aligned} _{more than Carl}^{Bianca has 19 marks} & \to b = c + 19 \\ _{(the total) is 153}^{The sum of the marks} & \to b + c = 153 \end{aligned}

$\begin{aligned} ^{\text{Bianca has } \text{19} \text{ marks}} _{\text{ more than Carl}} \quad & \rightarrow \quad b = c + \text{19} \\ \\ ^{\text{The sum of the marks}} _{\text{(the total) is } \text{153}} \quad & \rightarrow \quad b+c = \text{153} \end{aligned}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

\begin{aligned} b + c & = 153 \\ (c + 19) + c & = 153 \\ 2 c & = 153 - 19 \\ c & = \frac{134}{2} \\ = 67 \end{aligned}

$\begin{aligned} b+c &= \text{153} \\ (c+\text{19})+c &= \text{153} \\ 2c &= 153 - \text{19}\\ c &= \frac{\text{134}}{\text{2}}\\ &= \text{67} \end{aligned}$

This means that Carl's mark is $\text{67}$ .

STEP: Find Bianca's mark
[−1 point ⇒ 0 / 6 points left]

Now we can use Carl's mark and one of the equations we have to find Bianca's mark. The easiest way to do that is to substitute Carl's mark back into the first equation:

\begin{aligned} b & = c + 19 \\ = (67) + 19 \\ = 86 \end{aligned}

$\begin{aligned} b &= c+\text{19} \\ &= \left( \text{67} \right) + \text{19} \\ &= \text{86} \end{aligned}$

The students achieved these marks: Bianca earned $\text{86}$ and Carl earned $\text{67}$ .

Submit your answer as: and

Word problems: odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

The numbers are consecutive odd integers.
The sum of the numbers is $\text{24}$ .

What is the value of the smaller number?

Answer: The smaller number is .

numeric

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by choosing variables for the two numbers you are trying to find. For example, you can use

n_{1}

$n_{1}$ and

n_{2}

$n_{2}$ . Then write equations using these variables based on the facts given in the question.

STEP: Pick variables
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find one of them (the smaller one). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a sum of $\text{24}$ . We can solve this question using simultaneous equations.

\begin{aligned} _{we need to find}^{this is the number} \to n_{1} & = the smaller number \\ n_{2} & = the larger number \end{aligned}

$\begin{aligned} ^{\text{this is the number}}_{\text{we need to find}} \quad \rightarrow \quad n_{1} &= \text{the smaller number} \\ n_{2} &= \text{the larger number} \end{aligned}$

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

\begin{matrix} n_{2} = n_{1} + 2 \end{matrix}

$\begin{array}{c}n_{2} = n_{1} + 2\end{array}$

The " $\text{+2}$ " skips the even integer which sits between $n_{1}$ and $n_{2}$ .

The second fact about the numbers tells us that "the sum of the numbers is $\text{24}$ ". Remember that sum means addition. So:

\begin{matrix} n_{1} + n_{2} = 24 \end{matrix}

$\begin{array}{c}n_{1} + n_{2} = 24\end{array}$

STEP: Solve the equations for

n_{1}

$n_{1}$
[−2 points ⇒ 0 / 5 points left]

We can do this using substitution. If we substitute $n_2 = n_1 + 2$ into the equation $n_1 + n_2 = 24$ , the $n_2$ terms will disappear, leaving $n_1$ . Then we can solve for $n_1$ , which is exactly what we want.

\begin{aligned} n_{1} + (n_{1} + 2) & = 24 \\ 2 n_{1} + 2 & = 24 \\ 2 n_{1} & = 22 \\ n_{1} & = 11 \end{aligned}

$\begin{aligned} n_1 + \left(n_1 + 2\right) &= 24 \\ 2n_1 + 2 &= 24 \\ 2n_1 &= 22 \\ n_1 &= 11 \end{aligned}$

The result is $n_{1} = 11$ . Notice that this means that the other number, $n_{2}$ , must be $\text{13}$ , because $n_{2} = n_{1} + 2$ . This is perfect, because we know that the sum of the numbers is $\text{24}$ , and $11 + 13 = 24$ .

The smaller number is $\text{11}$ .

Submit your answer as:

Word problems: odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

The numbers are consecutive even integers.
The sum of the numbers is $\text{30}$ .

Determine the value of the smaller number.

Answer: The smaller number is .

numeric

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by choosing variables for the two numbers you are trying to find. For example, you can use

n_{1}

$n_{1}$ and

n_{2}

$n_{2}$ . Then write equations using these variables based on the facts given in the question.

STEP: Pick variables
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find one of them (the smaller one). We know certain things about these numbers: they are positive, they are consecutive even integers, and they have a sum of $\text{30}$ . We can solve this question using simultaneous equations.

\begin{aligned} _{we need to find}^{this is the number} \to n_{1} & = the smaller number \\ n_{2} & = the larger number \end{aligned}

$\begin{aligned} ^{\text{this is the number}}_{\text{we need to find}} \quad \rightarrow \quad n_{1} &= \text{the smaller number} \\ n_{2} &= \text{the larger number} \end{aligned}$

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact given in the question says that the numbers are "consecutive even integers". So both of the numbers are even, and they come one after another. For example, $\text{10}$ and $\text{12}$ are consecutive even numbers. Since we defined $n_{1}$ as the smaller number, $n_{2}$ must be $\text{2}$ more than $n_{1}$ .

\begin{matrix} n_{2} = n_{1} + 2 \end{matrix}

$\begin{array}{c}n_{2} = n_{1} + 2\end{array}$

The " $\text{+2}$ " skips the odd integer which sits between $n_{1}$ and $n_{2}$ .

The second fact about the numbers tells us that "the sum of the numbers is $\text{30}$ ". Remember that sum means addition. So:

\begin{matrix} n_{1} + n_{2} = 30 \end{matrix}

$\begin{array}{c}n_{1} + n_{2} = 30\end{array}$

STEP: Solve the equations for

n_{1}

$n_{1}$
[−2 points ⇒ 0 / 5 points left]

We can do this using substitution. If we substitute $n_2 = n_1 + 2$ into the equation $n_1 + n_2 = 30$ , the $n_2$ terms will disappear, leaving $n_1$ . Then we can solve for $n_1$ , which is exactly what we want.

\begin{aligned} n_{1} + (n_{1} + 2) & = 30 \\ 2 n_{1} + 2 & = 30 \\ 2 n_{1} & = 28 \\ n_{1} & = 14 \end{aligned}

$\begin{aligned} n_1 + \left(n_1 + 2\right) &= 30 \\ 2n_1 + 2 &= 30 \\ 2n_1 &= 28 \\ n_1 &= 14 \end{aligned}$

The result is $n_{1} = 14$ . Notice that this means that the other number, $n_{2}$ , must be $\text{16}$ , because $n_{2} = n_{1} + 2$ . This is perfect, because we know that the sum of the numbers is $\text{30}$ , and $14 + 16 = 30$ .

The smaller number is $\text{14}$ .

Submit your answer as:

Word problems: odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

The numbers are consecutive odd integers.
The sum of the numbers is $\text{32}$ .

What is the value of the larger number?

Answer: The larger number is .

numeric

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by choosing variables for the two numbers you are trying to find. For example, you can use

n_{1}

$n_{1}$ and

n_{2}

$n_{2}$ . Then write equations using these variables based on the facts given in the question.

STEP: Pick variables
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find one of them (the larger one). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a sum of $\text{32}$ . We can solve this question using simultaneous equations.

\begin{aligned} n_{1} & = the smaller number \\ _{we need to find}^{this is the number} \to n_{2} & = the larger number \end{aligned}

$\begin{aligned} n_{1} &= \text{the smaller number} \\ ^{\text{this is the number}}_{\text{we need to find}} \quad \rightarrow \quad n_{2} &= \text{the larger number} \end{aligned}$

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

\begin{matrix} n_{2} = n_{1} + 2 \end{matrix}

$\begin{array}{c}n_{2} = n_{1} + 2\end{array}$

The " $\text{+2}$ " skips the even integer which sits between $n_{1}$ and $n_{2}$ .

The second fact about the numbers tells us that "the sum of the numbers is $\text{32}$ ". Remember that sum means addition. So:

\begin{matrix} n_{1} + n_{2} = 32 \end{matrix}

$\begin{array}{c}n_{1} + n_{2} = 32\end{array}$

STEP: Solve the equations for

n_{2}

$n_{2}$
[−2 points ⇒ 0 / 5 points left]

\begin{aligned} n_{2} & = n_{1} + 2 \\ n_{2} - 2 & = n_{1} \end{aligned}

$\begin{aligned} n_2 &= n_1 + 2 \\ n_2 - 2 &= n_1 \end{aligned}$

Now substitute $n_2 - 2$ into the other equation and solve for $n_2$ .

\begin{aligned} (n_{2} - 2) + n_{2} & = 32 \\ 2 n_{2} - 2 & = 32 \\ 2 n_{2} & = 34 \\ n_{2} & = 17 \end{aligned}

$\begin{aligned} \left(n_2 - 2\right) + n_2 &= 32 \\ 2n_2 - 2 &= 32 \\ 2n_2 &= 34 \\ n_2 &= 17 \end{aligned}$

The result is $n_{2} = 17$ . Notice that this means that the other number, $n_{1}$ , must be $\text{15}$ , because $n_{2} = n_{1} + 2$ . This is perfect, because we know that the sum of the numbers is $\text{32}$ , and $17 + 15 = 32$ .

The larger number is $\text{17}$ .

Submit your answer as:

Word problems: checking answers

Balarabe is $\text{9}$ years younger than her sister, Nyakallo. In $\text{5}$ years, Nyakallo will be $\text{4}$ times as old as Balarabe. How old is Balarabe now?

INSTRUCTION: If there is no acceptable solution, type 'no solution' in the answer box.

Answer:

Balarabe is years old.

one-of

type(string.nocase)

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to find Balarabe's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.

STEP: Choose variables for each person's age
[−1 point ⇒ 4 / 5 points left]

We need to determine Balarabe's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Nyakallo and Balarabe. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

\begin{aligned} a_{N} & = Nyakallo's age \\ a_{B} & = Balarabe's age \end{aligned}

$\begin{aligned} a_{N} & = \text{Nyakallo's age} \\ a_{B} & = \text{Balarabe's age} \end{aligned}$

STEP: Write an equation based on the information in the question
[−1 point ⇒ 3 / 5 points left]

Now let's write equations using these variables. From the question we know that: "Balarabe is $\text{9}$ years younger than her sister, Nyakallo." We need to translate that into mathematics. The key word is younger, which tells us to use subtraction to relate the ages.

\begin{aligned} a_{B} = a_{N} - 9 \end{aligned}

$\begin{aligned} a_B = a_N - 9 \end{aligned}$

STEP: Write the equations
[−1 point ⇒ 2 / 5 points left]

We also know that: "In $\text{5}$ years, Nyakallo will be $\text{4}$ times as old as Balarabe." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

\begin{aligned} a_{N} + 5 & = Nyakallo's age in 5 years \\ a_{B} + 5 & = Balarabe's age in 5 years \end{aligned}

$\begin{aligned} a_{N} + 5 & = \text{Nyakallo's age in 5 years} \\ a_{B} + 5 & = \text{Balarabe's age in 5 years} \end{aligned}$

These are the ages at which Nyakallo will be $\text{4}$ times as old as Balarabe. We can put all this together as follows:

\begin{array}{rcl} _{in 5 years}^{Nyakallo's age} & = & 4 \times_{in 5 years}^{Balarabe's age} \\ ↓ \\ a_{N} + 5 & = & 4 (a_{B} + 5) \end{array}

$\begin{array}{rcl} ^{\text{Nyakallo's age}}_{\text{in 5 years}} & = & 4\; \times\; ^{\text{Balarabe's age}}_{\text{in 5 years}} \\ \quad & \downarrow & \quad \\ a_{N} + 5 & = & 4 \left( a_{B} + 5 \right) \end{array}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 0 / 5 points left]

Now we have to solve two equations, both of them including the variables $a_{N}$ and $a_{B}$ . The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is $a_B = a_N - 9$ . We can substitute this into the second equation and solve for $a_{N}$ .

\begin{aligned} a_{N} + 5 & = 4 (a_{B} + 5) \\ a_{N} + 5 & = 4 ((a_{N} - 9) + 5) \\ a_{N} + 5 & = 4 a_{N} - 16 \\ 21 & = 3 a_{N} \\ 7 & = a_{N} \end{aligned}

$\begin{aligned} a_{N} + 5 &= 4 \left(a_{B} + 5\right) \\ a_{N} + 5 &= 4 \left(\left(a_{N} - 9\right) + 5\right) \\ a_{N} + 5 &= 4 a_{N} - 16 \\ 21 &= 3 a_{N} \\ 7 &= a_{N} \end{aligned}$

Terrific: this means that Nyakallo is $\text{7}$ years old. But the question asked for us to find Balarabe's age. We can find it using the first equation, which relates the two ages:

\begin{aligned} a_{B} & = a_{N} - 9 \\ = (7) - 9 \\ = - 2 \end{aligned}

$\begin{aligned} a_B & = a_N - 9 \\ & = \left(7\right) - 9 \\ & = -2 \end{aligned}$

So we finally got the answer to the question. But wait a minute: $a_B$ represents the age of a person. It can't be negative! This either means that we made a mistake, or that there is no solution to the question. There is no mistake in the work: it turns out that the facts given about the two people's ages are not possible! The numbers agree with all the relationships given in the question, but we cannot forget that the numbers in this question have meaning. They refer to how many years it has been since someone was born. And that cannot be a negative number.

The correct answer is: no solution.

Submit your answer as:

Word problems: checking answers

Chinweike is $\text{12}$ years younger than her sister, Joanna. In $\text{8}$ years, Joanna will be $\text{2}$ times as old as Chinweike. How old is Chinweike now?

INSTRUCTION: If there is no acceptable solution, type 'no solution' in the answer box.

Answer:

Chinweike is years old.

numeric

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to find Chinweike's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.

STEP: Choose variables for each person's age
[−1 point ⇒ 4 / 5 points left]

We need to determine Chinweike's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Joanna and Chinweike. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

\begin{aligned} a_{J} & = Joanna's age \\ a_{C} & = Chinweike's age \end{aligned}

$\begin{aligned} a_{J} & = \text{Joanna's age} \\ a_{C} & = \text{Chinweike's age} \end{aligned}$

STEP: Write an equation based on the information in the question
[−1 point ⇒ 3 / 5 points left]

Now let's write equations using these variables. From the question we know that: "Chinweike is $\text{12}$ years younger than her sister, Joanna." We need to translate that into mathematics. The key word is younger, which tells us to use subtraction to relate the ages.

\begin{aligned} a_{C} = a_{J} - 12 \end{aligned}

$\begin{aligned} a_C = a_J - 12 \end{aligned}$

STEP: Write the equations
[−1 point ⇒ 2 / 5 points left]

We also know that: "In $\text{8}$ years, Joanna will be $\text{2}$ times as old as Chinweike." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

\begin{aligned} a_{J} + 8 & = Joanna's age in 8 years \\ a_{C} + 8 & = Chinweike's age in 8 years \end{aligned}

$\begin{aligned} a_{J} + 8 & = \text{Joanna's age in 8 years} \\ a_{C} + 8 & = \text{Chinweike's age in 8 years} \end{aligned}$

These are the ages at which Joanna will be $\text{2}$ times as old as Chinweike. We can put all this together as follows:

\begin{array}{rcl} _{in 8 years}^{Joanna's age} & = & 2 \times_{in 8 years}^{Chinweike's age} \\ ↓ \\ a_{J} + 8 & = & 2 (a_{C} + 8) \end{array}

$\begin{array}{rcl} ^{\text{Joanna's age}}_{\text{in 8 years}} & = & 2\; \times\; ^{\text{Chinweike's age}}_{\text{in 8 years}} \\ \quad & \downarrow & \quad \\ a_{J} + 8 & = & 2 \left( a_{C} + 8 \right) \end{array}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 0 / 5 points left]

Now we have to solve two equations, both of them including the variables $a_{J}$ and $a_{C}$ . The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is $a_C = a_J - 12$ . We can substitute this into the second equation and solve for $a_{J}$ .

\begin{aligned} a_{J} + 8 & = 2 (a_{C} + 8) \\ a_{J} + 8 & = 2 ((a_{J} - 12) + 8) \\ a_{J} + 8 & = 2 a_{J} - 8 \\ 16 & = a_{J} \end{aligned}

$\begin{aligned} a_{J} + 8 &= 2 \left(a_{C} + 8\right) \\ a_{J} + 8 &= 2 \left(\left(a_{J} - 12\right) + 8\right) \\ a_{J} + 8 &= 2 a_{J} - 8 \\ 16 &= a_{J} \\ \quad \end{aligned}$

Terrific: this means that Joanna is $\text{16}$ years old. But the question asked for us to find Chinweike's age. We can find it using the first equation, which relates the two ages:

\begin{aligned} a_{C} & = a_{J} - 12 \\ = (16) - 12 \\ = 4 \end{aligned}

$\begin{aligned} a_C & = a_J - 12 \\ & = \left(16\right) - 12 \\ & = 4 \end{aligned}$

So we finally got the answer to the question. Chinweike is $\text{4}$ years old.

The correct answer is: 4.

Submit your answer as:

Word problems: checking answers

Deborah is $\text{11}$ years younger than her sister, Qiniso. In $\text{9}$ years, Qiniso will be $\text{12}$ times as old as Deborah. How old is Deborah now?

INSTRUCTION: If there is no acceptable solution, type 'no solution' in the answer box.

Answer:

Deborah is years old.

one-of

type(string.nocase)

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You need to find Deborah's age based on the given information. The first thing to do is define variables for the two ages, which are unknown. Then write equations with those variables based on the facts in the question.

STEP: Choose variables for each person's age
[−1 point ⇒ 4 / 5 points left]

We need to determine Deborah's age based on the information given in the question. To do this, we can write equations to represent the information, and then solve the equations.

The first thing to do is define variables for the unknowns in the question. The unknown values are the ages of both Qiniso and Deborah. We can use any variables we want, but it is best to choose variables which represent the information we want. In this case, we want ages, and we need to distinguish them somehow. Here is one good pair of options:

\begin{aligned} a_{Q} & = Qiniso's age \\ a_{D} & = Deborah's age \end{aligned}

$\begin{aligned} a_{Q} & = \text{Qiniso's age} \\ a_{D} & = \text{Deborah's age} \end{aligned}$

STEP: Write an equation based on the information in the question
[−1 point ⇒ 3 / 5 points left]

Now let's write equations using these variables. From the question we know that: "Deborah is $\text{11}$ years younger than her sister, Qiniso." We need to translate that into mathematics. The key word is younger, which tells us to use subtraction to relate the ages.

\begin{aligned} a_{D} = a_{Q} - 11 \end{aligned}

$\begin{aligned} a_D = a_Q - 11 \end{aligned}$

STEP: Write the equations
[−1 point ⇒ 2 / 5 points left]

We also know that: "In $\text{9}$ years, Qiniso will be $\text{12}$ times as old as Deborah." Now we are looking into the future and we need to represent that information in maths. Using the variables we already have, we can write:

\begin{aligned} a_{Q} + 9 & = Qiniso's age in 9 years \\ a_{D} + 9 & = Deborah's age in 9 years \end{aligned}

$\begin{aligned} a_{Q} + 9 & = \text{Qiniso's age in 9 years} \\ a_{D} + 9 & = \text{Deborah's age in 9 years} \end{aligned}$

These are the ages at which Qiniso will be $\text{12}$ times as old as Deborah. We can put all this together as follows:

\begin{array}{rcl} _{in 9 years}^{Qiniso's age} & = & 12 \times_{in 9 years}^{Deborah's age} \\ ↓ \\ a_{Q} + 9 & = & 12 (a_{D} + 9) \end{array}

$\begin{array}{rcl} ^{\text{Qiniso's age}}_{\text{in 9 years}} & = & 12\; \times\; ^{\text{Deborah's age}}_{\text{in 9 years}} \\ \quad & \downarrow & \quad \\ a_{Q} + 9 & = & 12 \left( a_{D} + 9 \right) \end{array}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 0 / 5 points left]

Now we have to solve two equations, both of them including the variables $a_{Q}$ and $a_{D}$ . The easiest way to solve them is substitution (you can use elimination if you prefer). The first equation is $a_D = a_Q - 11$ . We can substitute this into the second equation and solve for $a_{Q}$ .

\begin{aligned} a_{Q} + 9 & = 12 (a_{D} + 9) \\ a_{Q} + 9 & = 12 ((a_{Q} - 11) + 9) \\ a_{Q} + 9 & = 12 a_{Q} - 24 \\ 33 & = 11 a_{Q} \\ 3 & = a_{Q} \end{aligned}

$\begin{aligned} a_{Q} + 9 &= 12 \left(a_{D} + 9\right) \\ a_{Q} + 9 &= 12 \left(\left(a_{Q} - 11\right) + 9\right) \\ a_{Q} + 9 &= 12 a_{Q} - 24 \\ 33 &= 11 a_{Q} \\ 3 &= a_{Q} \end{aligned}$

Terrific: this means that Qiniso is $\text{3}$ years old. But the question asked for us to find Deborah's age. We can find it using the first equation, which relates the two ages:

\begin{aligned} a_{D} & = a_{Q} - 11 \\ = (3) - 11 \\ = - 8 \end{aligned}

$\begin{aligned} a_D & = a_Q - 11 \\ & = \left(3\right) - 11 \\ & = -8 \end{aligned}$

So we finally got the answer to the question. But wait a minute: $a_D$ represents the age of a person. It can't be negative! This either means that we made a mistake, or that there is no solution to the question. There is no mistake in the work: it turns out that the facts given about the two people's ages are not possible! The numbers agree with all the relationships given in the question, but we cannot forget that the numbers in this question have meaning. They refer to how many years it has been since someone was born. And that cannot be a negative number.

The correct answer is: no solution.

Submit your answer as:

Word problems: products of odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

The numbers are consecutive even integers.
The product of the numbers is $\text{80}$ .

Find the value of the larger number.

Answer: The larger number is .

numeric

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by choosing variables for the two numbers you are trying to find. Then write equations using these variables based on the information given in the question.

STEP: Pick variables for the numbers
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find the larger number (not both). We know certain things about these numbers: they are positive, they are consecutive even integers, and they have a product of $\text{80}$ . We can use that information to solve this question using simultaneous equations.

\begin{aligned} n_{1} & = the smaller number \\ _{we need to find}^{this is the number} \to n_{2} & = the larger number \end{aligned}

$\begin{aligned} n_{1} &= \text{the smaller number} \\ ^{\text{this is the number}}_{\text{we need to find}} \quad \rightarrow \quad n_{2} &= \text{the larger number} \end{aligned}$

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

\begin{matrix} n_{2} = n_{1} + 2 \end{matrix}

$\begin{array}{c}n_{2} = n_{1} + 2\end{array}$

The " $\text{+2}$ " skips the odd integer which sits between $n_{1}$ and $n_{2}$ .

The second fact about the numbers tells us that "the product of the numbers is $\text{80}$ ". Remember that product means multiplication. So:

\begin{matrix} n_{1} \cdot n_{2} = 80 \end{matrix}

$\begin{array}{c}n_{1} \cdot n_{2} = 80\end{array}$

STEP: Solve the equations for

n_{2}

$n_{2}$
[−2 points ⇒ 0 / 5 points left]

Remember that we want the value of $n_2$ (the larger number). We can start by rearranging the first equation to make $n_1$ the subject. Then the substitution step will remove $n_1$ from the second equation and we can solve for $n_2$ . (This is not required - it just makes the solution faster.)

\begin{aligned} n_{2} & = n_{1} + 2 \\ n_{2} - 2 & = n_{1} \end{aligned}

$\begin{aligned} n_2 &= n_1 + 2 \\ n_2 - 2 &= n_1 \end{aligned}$

Now substitute $n_2 - 2$ into the other equation and solve for $n_2$ .

\begin{aligned} (n_{2} - 2) \cdot n_{2} & = 80 \\ n_{2}^{2} - 2 n_{2} & = 80 \\ n_{2}^{2} - 2 n_{2} - 80 & = 0 \\ (n_{2} - 10) (n_{2} + 8) & = 0 \end{aligned}

$\begin{aligned} \left(n_2 - 2\right) \cdot n_2 &= 80 \\ n_2^{\; 2} - 2n_2 &= 80 \\ n_2^{\; 2} - 2n_2 - 80 &= 0 \\ \left(n_2 - 10\right) \left(n_2 + 8\right) &= 0 \end{aligned}$

\begin{aligned} n_{2} = 10 and n_{2} = - 8 \end{aligned}

$\begin{aligned} n_2 = 10 \text{ and } n_2 = -8 \end{aligned}$

This solution led to two answers for $n_{2}$ . However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves $n_{2} = 10$ .

This means the other number, $n_{1}$ , must be $\text{8}$ , because $n_{2} = n_{1} + 2$ . This is perfect, because we also know that the product of the numbers is $\text{80}$ , and $10 \cdot 8 = 80$ .

The larger number is $\text{10}$ .

Submit your answer as:

Word problems: products of odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

The numbers are consecutive odd integers.
The product of the numbers is $\text{63}$ .

Find the value of the larger number.

Answer: The larger number is .

numeric

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by choosing variables for the two numbers you are trying to find. Then write equations using these variables based on the information given in the question.

STEP: Pick variables for the numbers
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find the larger number (not both). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a product of $\text{63}$ . We can use that information to solve this question using simultaneous equations.

\begin{aligned} n_{1} & = the smaller number \\ _{we need to find}^{this is the number} \to n_{2} & = the larger number \end{aligned}

$\begin{aligned} n_{1} &= \text{the smaller number} \\ ^{\text{this is the number}}_{\text{we need to find}} \quad \rightarrow \quad n_{2} &= \text{the larger number} \end{aligned}$

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

The first fact in the question says that the numbers are "consecutive odd integers". So both of the numbers are odd, and they come one after another. For example, $\text{11}$ and $\text{13}$ are consecutive odd numbers. Since we defined $n_{1}$ as the smaller number, $n_{2}$ must be $\text{2}$ more than $n_{1}$ .

\begin{matrix} n_{2} = n_{1} + 2 \end{matrix}

$\begin{array}{c}n_{2} = n_{1} + 2\end{array}$

The " $\text{+2}$ " skips the even integer which sits between $n_{1}$ and $n_{2}$ .

The second fact about the numbers tells us that "the product of the numbers is $\text{63}$ ". Remember that product means multiplication. So:

\begin{matrix} n_{1} \cdot n_{2} = 63 \end{matrix}

$\begin{array}{c}n_{1} \cdot n_{2} = 63\end{array}$

STEP: Solve the equations for

n_{2}

$n_{2}$
[−2 points ⇒ 0 / 5 points left]

\begin{aligned} n_{2} & = n_{1} + 2 \\ n_{2} - 2 & = n_{1} \end{aligned}

$\begin{aligned} n_2 &= n_1 + 2 \\ n_2 - 2 &= n_1 \end{aligned}$

Now substitute $n_2 - 2$ into the other equation and solve for $n_2$ .

\begin{aligned} (n_{2} - 2) \cdot n_{2} & = 63 \\ n_{2}^{2} - 2 n_{2} & = 63 \\ n_{2}^{2} - 2 n_{2} - 63 & = 0 \\ (n_{2} - 9) (n_{2} + 7) & = 0 \end{aligned}

$\begin{aligned} \left(n_2 - 2\right) \cdot n_2 &= 63 \\ n_2^{\; 2} - 2n_2 &= 63 \\ n_2^{\; 2} - 2n_2 - 63 &= 0 \\ \left(n_2 - 9\right) \left(n_2 + 7\right) &= 0 \end{aligned}$

\begin{aligned} n_{2} = 9 and n_{2} = - 7 \end{aligned}

$\begin{aligned} n_2 = 9 \text{ and } n_2 = -7 \end{aligned}$

This solution led to two answers for $n_{2}$ . However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves $n_{2} = 9$ .

This means the other number, $n_{1}$ , must be $\text{7}$ , because $n_{2} = n_{1} + 2$ . This is perfect, because we also know that the product of the numbers is $\text{63}$ , and $9 \cdot 7 = 63$ .

The larger number is $\text{9}$ .

Submit your answer as:

Word problems: products of odd and even numbers

This question is about two positive numbers. Here are facts about these numbers:

The numbers are consecutive odd integers.
The product of the numbers is $\text{3}$ .

What is the value of the smaller number?

Answer: The smaller number is .

numeric

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by choosing variables for the two numbers you are trying to find. Then write equations using these variables based on the information given in the question.

STEP: Pick variables for the numbers
[−1 point ⇒ 4 / 5 points left]

This question is about two unknown numbers. And we need to find the smaller number (not both). We know certain things about these numbers: they are positive, they are consecutive odd integers, and they have a product of $\text{3}$ . We can use that information to solve this question using simultaneous equations.

\begin{aligned} _{we need to find}^{this is the number} \to n_{1} & = the smaller number \\ n_{2} & = the larger number \end{aligned}

$\begin{aligned} ^{\text{this is the number}}_{\text{we need to find}} \quad \rightarrow \quad n_{1} &= \text{the smaller number} \\ n_{2} &= \text{the larger number} \end{aligned}$

STEP: Write two equations
[−2 points ⇒ 2 / 5 points left]

\begin{matrix} n_{2} = n_{1} + 2 \end{matrix}

$\begin{array}{c}n_{2} = n_{1} + 2\end{array}$

The " $\text{+2}$ " skips the even integer which sits between $n_{1}$ and $n_{2}$ .

The second fact about the numbers tells us that "the product of the numbers is $\text{3}$ ". Remember that product means multiplication. So:

\begin{matrix} n_{1} \cdot n_{2} = 3 \end{matrix}

$\begin{array}{c}n_{1} \cdot n_{2} = 3\end{array}$

STEP: Solve the equations for

n_{1}

$n_{1}$
[−2 points ⇒ 0 / 5 points left]

If we subsitute $n_2 = n_1 + 2$ into the equation $n_1 \cdot n_2 = 3$ , the $n_2$ terms will disappear, leaving $n_1$ . Then we can solve for $n_1$ , which is exactly what we want.

\begin{aligned} n_{1} \cdot (n_{1} + 2) & = 3 \\ n_{1}^{2} + 2 n_{1} & = 3 \\ n_{1}^{2} + 2 n_{1} - 3 & = 0 \\ (n_{1} + 3) (n_{1} - 1) & = 0 \end{aligned}

$\begin{aligned} n_1 \cdot \left(n_1 + 2\right) &= 3 \\ n_1^{\; 2} + 2n_1 &= 3 \\ n_1^{\; 2} + 2n_1 - 3 &= 0 \\ \left(n_1 + 3\right) \left(n_1 - 1\right) &= 0 \end{aligned}$

\begin{aligned} n_{1} = - 3 and n_{1} = 1 \end{aligned}

$\begin{aligned} n_1 = -3 \quad \text{and} \quad n_1 = 1 \end{aligned}$

This solution led to two answers for $n_{1}$ . However, remember that the numbers in this question are positive. So we can throw away the negative answer, which leaves $n_{1} = 1$ .

This means the other number, $n_{2}$ , must be $\text{3}$ , because $n_{2} = n_{1} + 2$ . This is perfect, because we also know that the product of the numbers is $\text{3}$ , and $1 \cdot 3 = 3$ .

The smaller number is $\text{1}$ .

Submit your answer as:

Word problems: finding consecutive numbers

Here are some facts about two numbers:

the numbers are consecutive integers
both numbers are negative
the product of the numbers is $\text{30}$

What are the two numbers?

TIP: 'Product' means multiplication; 'consecutive' means 'following each other without gaps between.'

INSTRUCTION: Write your answers in the boxes below. It does not matter what order your answers are in.

Answer: The numbers are and .

numeric

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You should represent the numbers in the question by $n$ and $n+1$ . Write an equation based on the fact that the product of those two numbers is $\text{30}$ .

STEP: Pick variables to represent the numbers
[−1 point ⇒ 5 / 6 points left]

This question asks us to find two numbers which have a product of $\text{30}$ . That means if we multiply the numbers we get $\text{30}$ .

\begin{aligned} n_{1} & = the first number \\ n_{2} & = the next number \end{aligned}

$\begin{aligned} n_{1} & = \text{the first number} \\ n_{2} & = \text{the next number} \end{aligned}$

STEP: Write an equation based on the information in the question
[−1 point ⇒ 4 / 6 points left]

The question tells us that the product of these numbers is $\text{30}$ . In other words, if we multiply the numbers, we get $\text{30}$ . As an equation this is:

\begin{matrix} n_{1} \cdot n_{2} = 30 \end{matrix}

$\begin{array}{c}n_{1} \cdot n_{2} = 30\end{array}$

STEP: Connect the two variables
[−1 point ⇒ 3 / 6 points left]

Now we can use the fact that the numbers are consecutive. So if $n_{1}$ represents the first number, then the next number must be one more than $n_{1}$ .

\begin{aligned} If: & n_{2} = the number following n_{1} \\ then: & n_{2} = n_{1} + 1 \end{aligned}

$\begin{aligned} \text{If:} \qquad & n_{2} = \text{the number following } n_{1} \\ \\ \text{then:} \qquad & n_{2} = n_{1} + 1 \end{aligned}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have two equations about the two numbers.

\begin{aligned} n_{1} \cdot n_{2} = 30 \\ n_{2} = n_{1} + 1 \end{aligned}

$\begin{aligned} & n_{1} \cdot n_{2} = 30 \\ & n_{2} = n_{1} + 1 \end{aligned}$

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for $n_{1}$ .

\begin{aligned} n_{1} \cdot (n_{1} + 1) & = 30 \\ (n_{1})^{2} + n_{1} - 30 & = 0 \\ (n_{1} + 6) (n_{1} - 5) & = 0 \end{aligned}

$\begin{aligned} n_{1} \cdot \left(n_{1} + 1 \right) & = 30 \\ (n_{1})^2 + n_{1} - 30 &= 0 \\ \left( n_{1} + 6\right) \left(n_{1}-5 \right) &= 0 \end{aligned}$

\begin{aligned} n_{1} = - 6 and n_{1} = 5 \end{aligned}

$\begin{aligned} n_{1} = -6 \quad \text{and} \quad n_{1} = 5 \end{aligned}$

The solutions to the equation are $n_{1} = -6$ or $n_{1} = 5$ . This means the first number is either $\text{−6}$ or $\text{5}$ .

STEP: Find the final answers
[−1 point ⇒ 0 / 6 points left]

Wait a minute! The numbers $\text{−6}$ and $\text{5}$ are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are negative. That means we must throw out the $n_{1} = 5$ , so the first number is $n_{1} = -6$ .

Based on that we can find the second number. The second number is $n_{1} + 1$ , which is equal to $\left(-6\right) + 1 = -5$ .

The two consecutive integers are $\text{−6}$ and $\text{−5}$ .

Submit your answer as: and

Word problems: finding consecutive numbers

Here are some facts about two numbers:

the numbers are consecutive integers
both numbers are positive
the product of the numbers is $\text{110}$

What are the two numbers?

TIP: 'Product' means multiplication; 'consecutive' means 'following each other without gaps between.'

INSTRUCTION: Write your answers in the boxes below. It does not matter what order your answers are in.

Answer: The numbers are and .

numeric

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You should represent the numbers in the question by $n$ and $n+1$ . Write an equation based on the fact that the product of those two numbers is $\text{110}$ .

STEP: Pick variables to represent the numbers
[−1 point ⇒ 5 / 6 points left]

This question asks us to find two numbers which have a product of $\text{110}$ . That means if we multiply the numbers we get $\text{110}$ .

\begin{aligned} n_{1} & = the first number \\ n_{2} & = the next number \end{aligned}

$\begin{aligned} n_{1} & = \text{the first number} \\ n_{2} & = \text{the next number} \end{aligned}$

STEP: Write an equation based on the information in the question
[−1 point ⇒ 4 / 6 points left]

The question tells us that the product of these numbers is $\text{110}$ . In other words, if we multiply the numbers, we get $\text{110}$ . As an equation this is:

\begin{matrix} n_{1} \cdot n_{2} = 110 \end{matrix}

$\begin{array}{c}n_{1} \cdot n_{2} = 110\end{array}$

STEP: Connect the two variables
[−1 point ⇒ 3 / 6 points left]

Now we can use the fact that the numbers are consecutive. So if $n_{1}$ represents the first number, then the next number must be one more than $n_{1}$ .

\begin{aligned} If: & n_{2} = the number following n_{1} \\ then: & n_{2} = n_{1} + 1 \end{aligned}

$\begin{aligned} \text{If:} \qquad & n_{2} = \text{the number following } n_{1} \\ \\ \text{then:} \qquad & n_{2} = n_{1} + 1 \end{aligned}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have two equations about the two numbers.

\begin{aligned} n_{1} \cdot n_{2} = 110 \\ n_{2} = n_{1} + 1 \end{aligned}

$\begin{aligned} & n_{1} \cdot n_{2} = 110 \\ & n_{2} = n_{1} + 1 \end{aligned}$

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for $n_{1}$ .

\begin{aligned} n_{1} \cdot (n_{1} + 1) & = 110 \\ (n_{1})^{2} + n_{1} - 110 & = 0 \\ (n_{1} + 11) (n_{1} - 10) & = 0 \end{aligned}

$\begin{aligned} n_{1} \cdot \left(n_{1} + 1 \right) & = 110 \\ (n_{1})^2 + n_{1} - 110 &= 0 \\ \left( n_{1} + 11\right) \left(n_{1}-10 \right) &= 0 \end{aligned}$

\begin{aligned} n_{1} = - 11 and n_{1} = 10 \end{aligned}

$\begin{aligned} n_{1} = -11 \quad \text{and} \quad n_{1} = 10 \end{aligned}$

The solutions to the equation are $n_{1} = -11$ or $n_{1} = 10$ . This means the first number is either $\text{−11}$ or $\text{10}$ .

STEP: Find the final answers
[−1 point ⇒ 0 / 6 points left]

Wait a minute! The numbers $\text{−11}$ and $\text{10}$ are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are positive. That means we must throw out the $n_{1} = -11$ , so the first number is $n_{1} = 10$ .

Based on that we can find the second number. The second number is $n_{1} + 1$ , which is equal to $\left(10\right) + 1 = 11$ .

The two consecutive integers are $\text{10}$ and $\text{11}$ .

Submit your answer as: and

Word problems: finding consecutive numbers

Here are some facts about two numbers:

the numbers are consecutive integers
both numbers are negative
the product of the numbers is $\text{306}$

What are the two numbers?

TIP: 'Product' means multiplication; 'consecutive' means 'following each other without gaps between.'

INSTRUCTION: Write your answers in the boxes below. It does not matter what order your answers are in.

Answer: The numbers are and .

numeric

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

You should represent the numbers in the question by $n$ and $n+1$ . Write an equation based on the fact that the product of those two numbers is $\text{306}$ .

STEP: Pick variables to represent the numbers
[−1 point ⇒ 5 / 6 points left]

This question asks us to find two numbers which have a product of $\text{306}$ . That means if we multiply the numbers we get $\text{306}$ .

\begin{aligned} n_{1} & = the first number \\ n_{2} & = the next number \end{aligned}

$\begin{aligned} n_{1} & = \text{the first number} \\ n_{2} & = \text{the next number} \end{aligned}$

STEP: Write an equation based on the information in the question
[−1 point ⇒ 4 / 6 points left]

The question tells us that the product of these numbers is $\text{306}$ . In other words, if we multiply the numbers, we get $\text{306}$ . As an equation this is:

\begin{matrix} n_{1} \cdot n_{2} = 306 \end{matrix}

$\begin{array}{c}n_{1} \cdot n_{2} = 306\end{array}$

STEP: Connect the two variables
[−1 point ⇒ 3 / 6 points left]

Now we can use the fact that the numbers are consecutive. So if $n_{1}$ represents the first number, then the next number must be one more than $n_{1}$ .

\begin{aligned} If: & n_{2} = the number following n_{1} \\ then: & n_{2} = n_{1} + 1 \end{aligned}

$\begin{aligned} \text{If:} \qquad & n_{2} = \text{the number following } n_{1} \\ \\ \text{then:} \qquad & n_{2} = n_{1} + 1 \end{aligned}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

Now we have two equations about the two numbers.

\begin{aligned} n_{1} \cdot n_{2} = 306 \\ n_{2} = n_{1} + 1 \end{aligned}

$\begin{aligned} & n_{1} \cdot n_{2} = 306 \\ & n_{2} = n_{1} + 1 \end{aligned}$

We can solve these equations simultaneously. Substitute the second equation into the first. Then solve for $n_{1}$ .

\begin{aligned} n_{1} \cdot (n_{1} + 1) & = 306 \\ (n_{1})^{2} + n_{1} - 306 & = 0 \\ (n_{1} + 18) (n_{1} - 17) & = 0 \end{aligned}

$\begin{aligned} n_{1} \cdot \left(n_{1} + 1 \right) & = 306 \\ (n_{1})^2 + n_{1} - 306 &= 0 \\ \left( n_{1} + 18\right) \left(n_{1}-17 \right) &= 0 \end{aligned}$

\begin{aligned} n_{1} = - 18 and n_{1} = 17 \end{aligned}

$\begin{aligned} n_{1} = -18 \quad \text{and} \quad n_{1} = 17 \end{aligned}$

The solutions to the equation are $n_{1} = -18$ or $n_{1} = 17$ . This means the first number is either $\text{−18}$ or $\text{17}$ .

STEP: Find the final answers
[−1 point ⇒ 0 / 6 points left]

Wait a minute! The numbers $\text{−18}$ and $\text{17}$ are not consecutive... and how did we get two answers for the first number? What's going on?

Remember that the question says that both of the numbers are negative. That means we must throw out the $n_{1} = 17$ , so the first number is $n_{1} = -18$ .

Based on that we can find the second number. The second number is $n_{1} + 1$ , which is equal to $\left(-18\right) + 1 = -17$ .

The two consecutive integers are $\text{−18}$ and $\text{−17}$ .

Submit your answer as: and

Word problems: an age-old question

Jacob has a son, Chinedu. Here are some facts about how old Jacob and Chinedu are:

Jacob is $\text{9}$ times as old as Chinedu right now.
$\text{5}$ years from now, Jacob will be $\text{4}$ times as old as Chinedu.

How old are Jacob and Chinedu now?

Answer:

Jacob is years old and Chinedu is years old.

numeric

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.

STEP: Choose variables for the information
[−1 point ⇒ 5 / 6 points left]

In this question we want to find the age of two people: Jacob and his son, Chinedu. We can solve this by setting up two equations and solving them simultaneously.

\begin{aligned} Let j = Jacob's age \\ Let c = Chinedu's age \end{aligned}

$\begin{aligned} & \text{Let } j = \text{Jacob's age} \\ & \text{Let } c = \text{Chinedu's age} \end{aligned}$

STEP: Write equations based on the information in the question
[−2 points ⇒ 3 / 6 points left]

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Jacob is now $\text{9}$ times as old as Chinedu." As an equation, this is:

\begin{matrix} Ages now: j = 9 c \end{matrix}

$\begin{array}{c}\text{Ages now: } \quad j = \text{9} c\end{array}$

The second piece of information says that in " $\text{5}$ years... Jacob will be $\text{4}$ times as old as his son." In $\text{5}$ years Jacob will be $j + \text{5}$ years old, and similarly Chinedu will be $c + \text{5}$ years old. Then:

\begin{matrix} Ages in 5 years: j + 5 = 4 (c + 5) \end{matrix}

$\begin{array}{c}\text{Ages in } \text{5} \text{ years: } \quad j + \text{5} = \text{4}(c+\text{5})\end{array}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

\begin{aligned} j + 5 & = 4 (c + 5) \\ (9 c) + 5 & = 4 c + 20 \\ 5 c & = 15 \\ c & = 3 \end{aligned}

$\begin{aligned} j + \text{5} &= \text{4}(c+\text{5}) \\ (\text{9}c) + \text{5} &= \text{4}c+\text{20} \\ 5 c &= \text{15} \\ c & = \text{3} \end{aligned}$

Great! Now we know that Chinedu is $\text{3}$ years old.

STEP: Use Chinedu's age to find Jacob's age
[−1 point ⇒ 0 / 6 points left]

We can now find Jacob's age. Substitute Chinedu's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

\begin{aligned} j & = 9 c \\ = 9 (3) \\ = 27 \end{aligned}

$\begin{aligned} j &= \text{9}c \\ &= \text{9} \left( \text{3} \right) \\ &= \text{27} \end{aligned}$

Write your final answer: Jacob is $\text{27}$ years old and Chinedu is $\text{3}$ years old.

Submit your answer as: and

Word problems: an age-old question

Ben has a son, Gift. Here are some facts about how old Ben and Gift are:

Ben is $\text{11}$ times as old as Gift right now.
$\text{4}$ years from now, Ben will be $\text{6}$ times as old as Gift.

How old are Ben and Gift now?

Answer:

Ben is years old and Gift is years old.

numeric

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.

STEP: Choose variables for the information
[−1 point ⇒ 5 / 6 points left]

In this question we want to find the age of two people: Ben and his son, Gift. We can solve this by setting up two equations and solving them simultaneously.

\begin{aligned} Let b = Ben's age \\ Let g = Gift's age \end{aligned}

$\begin{aligned} & \text{Let } b = \text{Ben's age} \\ & \text{Let } g = \text{Gift's age} \end{aligned}$

STEP: Write equations based on the information in the question
[−2 points ⇒ 3 / 6 points left]

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Ben is now $\text{11}$ times as old as Gift." As an equation, this is:

\begin{matrix} Ages now: b = 11 g \end{matrix}

$\begin{array}{c}\text{Ages now: } \quad b = \text{11} g\end{array}$

The second piece of information says that in " $\text{4}$ years... Ben will be $\text{6}$ times as old as his son." In $\text{4}$ years Ben will be $b + \text{4}$ years old, and similarly Gift will be $g + \text{4}$ years old. Then:

\begin{matrix} Ages in 4 years: b + 4 = 6 (g + 4) \end{matrix}

$\begin{array}{c}\text{Ages in } \text{4} \text{ years: } \quad b + \text{4} = \text{6}(g+\text{4})\end{array}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

\begin{aligned} b + 4 & = 6 (g + 4) \\ (11 g) + 4 & = 6 g + 24 \\ 5 g & = 20 \\ g & = 4 \end{aligned}

$\begin{aligned} b + \text{4} &= \text{6}(g+\text{4}) \\ (\text{11}g) + \text{4} &= \text{6}g+\text{24} \\ 5 g &= \text{20} \\ g & = \text{4} \end{aligned}$

Great! Now we know that Gift is $\text{4}$ years old.

STEP: Use Gift's age to find Ben's age
[−1 point ⇒ 0 / 6 points left]

We can now find Ben's age. Substitute Gift's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

\begin{aligned} b & = 11 g \\ = 11 (4) \\ = 44 \end{aligned}

$\begin{aligned} b &= \text{11}g \\ &= \text{11} \left( \text{4} \right) \\ &= \text{44} \end{aligned}$

Write your final answer: Ben is $\text{44}$ years old and Gift is $\text{4}$ years old.

Submit your answer as: and

Word problems: an age-old question

Sehlolo has a son, Peter. Here are some facts about how old Sehlolo and Peter are:

Sehlolo is $\text{9}$ times as old as Peter right now.
$\text{9}$ years from now, Sehlolo will be $\text{3}$ times as old as Peter.

How old are Sehlolo and Peter now?

Answer:

Sehlolo is years old and Peter is years old.

numeric

HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

Start by choosing variables for the things you want to know. Then write equations with those variables to summarise the information in the question.

STEP: Choose variables for the information
[−1 point ⇒ 5 / 6 points left]

In this question we want to find the age of two people: Sehlolo and his son, Peter. We can solve this by setting up two equations and solving them simultaneously.

\begin{aligned} Let s = Sehlolo's age \\ Let p = Peter's age \end{aligned}

$\begin{aligned} & \text{Let } s = \text{Sehlolo's age} \\ & \text{Let } p = \text{Peter's age} \end{aligned}$

STEP: Write equations based on the information in the question
[−2 points ⇒ 3 / 6 points left]

Now we want to use those variables to write equations. The first piece of information from the question tells us that "Sehlolo is now $\text{9}$ times as old as Peter." As an equation, this is:

\begin{matrix} Ages now: s = 9 p \end{matrix}

$\begin{array}{c}\text{Ages now: } \quad s = \text{9} p\end{array}$

The second piece of information says that in " $\text{9}$ years... Sehlolo will be $\text{3}$ times as old as his son." In $\text{9}$ years Sehlolo will be $s + \text{9}$ years old, and similarly Peter will be $p + \text{9}$ years old. Then:

\begin{matrix} Ages in 9 years: s + 9 = 3 (p + 9) \end{matrix}

$\begin{array}{c}\text{Ages in } \text{9} \text{ years: } \quad s + \text{9} = \text{3}(p+\text{9})\end{array}$

STEP: Solve the equations simultaneously
[−2 points ⇒ 1 / 6 points left]

\begin{aligned} s + 9 & = 3 (p + 9) \\ (9 p) + 9 & = 3 p + 27 \\ 6 p & = 18 \\ p & = 3 \end{aligned}

$\begin{aligned} s + \text{9} &= \text{3}(p+\text{9}) \\ (\text{9}p) + \text{9} &= \text{3}p+\text{27} \\ 6 p &= \text{18} \\ p & = \text{3} \end{aligned}$

Great! Now we know that Peter is $\text{3}$ years old.

STEP: Use Peter's age to find Sehlolo's age
[−1 point ⇒ 0 / 6 points left]

We can now find Sehlolo's age. Substitute Peter's age into one of the equations to do this. In this case, the first equation is the simpler choice for this calculation.

\begin{aligned} s & = 9 p \\ = 9 (3) \\ = 27 \end{aligned}

$\begin{aligned} s &= \text{9}p \\ &= \text{9} \left( \text{3} \right) \\ &= \text{27} \end{aligned}$

Write your final answer: Sehlolo is $\text{27}$ years old and Peter is $\text{3}$ years old.

Submit your answer as: and

Word Problems: checking answers

Suppose you must solve this word problem:

A marathon is a long-distance running race with an official distance of $\text{42,195}\ \text{kilometres}$ . $\text{1 000}$ people ran the "It Ain't Over Till It's Over" marathon in Port Elizabeth. The second-place runner finished $\text{19,82}$ seconds after the winner. The third runner was $\text{16,64}$ seconds behind the second runner, and the fourth runner was $\text{24,21}$ seconds behind the third runner. If the second-place runner had an average speed of $\text{18,72}\ \text{km/h}$ , what was the winner's average speed for the marathon?

Can the answer to the question above be a negative number?

Answer:

Can the answer be negative?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

STEP: Decide if the answer can be negative
[−1 point ⇒ 0 / 1 points left]

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a negative number or not.

Focus on what the problem asks: what was the winner's average speed for the marathon. That means we need a speed value. Negative speed is not possible: it must be a positive number.

The correct answer is: No, it cannot be negative.

Submit your answer as:

Word Problems: checking answers

Suppose you must solve this word problem:

A marathon is a long-distance running race with an official distance of $\text{42,195}\ \text{kilometres}$ . $\text{1 000}$ people ran the "One Metre at a Time" marathon in Polokwane. The second-place runner finished $\text{15,69}$ seconds after the winner. The third runner was $\text{12,51}$ seconds behind the second runner, and the fourth runner was $\text{29,41}$ seconds behind the third runner. If the second-place runner had an average speed of $\text{18,57}\ \text{km/h}$ , how many of the runners took more than $\text{2,36}$ hours to finish the marathon?

Can the answer to the question above be a negative number?

Answer:

Can the answer be negative?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

STEP: Decide if the answer can be negative
[−1 point ⇒ 0 / 1 points left]

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a negative number or not.

Focus on what the problem asks: how many of the runners took more than $\text{2,36}$ hours to finish the marathon. So the question is about the number of people who finished the race after $\text{2,36}$ hours had passed. The answer cannot be negative. The number of people to finish after $\text{2,36}$ hours might be $\text{3}$ or $\text{7}$ people, but it cannot be $-\text{5}$ people. In other words, the number of people must be a positive number or zero, but it cannot be negative.

The correct answer is: No, it cannot be negative.

Submit your answer as:

Word Problems: checking answers

Suppose you must solve this word problem:

A marathon is a long-distance running race with an official distance of $\text{42,195}\ \text{kilometres}$ . $\text{1 000}$ people ran the "One Metre at a Time" marathon in Port Elizabeth. The second-place runner finished $\text{15,46}$ seconds after the winner. The third runner was $\text{14,66}$ seconds behind the second runner, and the fourth runner was $\text{27,92}$ seconds behind the third runner. If the second-place runner had an average speed of $\text{19,93}\ \text{km/h}$ , how many of these runners finished the marathon in less than $\text{2,18}$ hours?

Can the answer to the question above be a negative number?

Answer:

Can the answer be negative?

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

STEP: Decide if the answer can be negative
[−1 point ⇒ 0 / 1 points left]

For this question, we do not have to solve the word problem about the marathon. We only need to decide if the answer to that question can be a negative number or not.

Focus on what the problem asks: how many of these runners finished the marathon in less than $\text{2,18}$ hours. So the question is about the number of people who finished the race before $\text{2,18}$ hours had passed. The answer cannot be negative. The number of people to finish before $\text{2,18}$ hours might be $\text{3}$ or $\text{7}$ people, but it cannot be $-\text{5}$ people. In other words, the number of people must be a positive number or zero, but it cannot be negative.

The correct answer is: No, it cannot be negative.

Submit your answer as:

Word problems: rectangle facts

The diagonal of a rectangle is $\text{32} \ \text{cm}$ more than its width. The length of the same rectangle is $\text{16}\ \text{cm}$ more than its width.

What are the width and the length of the rectangle?

Answer: The width is

cm

$\text{cm}$ and length is

cm

$\text{cm}$ .

numeric

HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.

STEP: Draw a diagram and label it
[−1 point ⇒ 7 / 8 points left]

Notice we have added labels for the three parts of the rectangle mentioned in the question: $d$ is the diagonal, $w$ is the width, and $l$ is the length.

STEP: Connect the variables in the diagram
[−2 points ⇒ 5 / 8 points left]

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

\begin{array}{lcl} _{32 cm more than its width}^{the diagonal of the rectangle is} & \to d & = w + 32 \\ _{16 cm more than its width}^{the length of the rectangle is} & \to l & = w + 16 \end{array}

$\begin{array}{lcl} ^{\text{the diagonal of the rectangle is}}_{ \text{32} \ \text{cm} \text{ more than its width}} & \rightarrow \quad d &= w + \text{32} \\ \\ ^{\text{the length of the rectangle is}}_{\text{16}\ \text{cm} \text{ more than its width}} & \rightarrow \quad l &= w + \text{16} \end{array}$

STEP: Use the Theorem of Pythagoras and solve the equations simultaneously
[−2 points ⇒ 3 / 8 points left]

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

\begin{aligned} d^{2} & = w^{2} + l^{2} \end{aligned}

$\begin{aligned} d^2 &= w^2 + l^2 \end{aligned}$

And now we can substitute in the equations from above. Specifically we can substitute $w + \text{32}$ in for $d$ and $w + \text{16}$ in for $l$ . Then expand the binomials and simplify the equation as much as possible.

\begin{aligned} d^{2} & = w^{2} + l^{2} \\ (w + 32)^{2} & = w^{2} + (w + 16)^{2} \\ w^{2} + 64 w + 1 024 & = w^{2} + (w^{2} + 32 w + 256) \\ 0 & = w^{2} - 32 w - 768 \end{aligned}

$\begin{aligned} d^2 &= w^2 + l^2 \\ (w + \text{32})^2 & = w^2 + (w + \text{16})^2\\ w^2 + \text{64}w + \text{1 024} & = w^2 + (w^2 + \text{32}w + \text{256})\\ 0 & = w^2 - \text{32}w - \text{768} \end{aligned}$

STEP: Solve the equation for

w

$w$
[−2 points ⇒ 1 / 8 points left]

We have a quadratic equation in standard form! It is time to solve it.

\begin{aligned} 0 & = w^{2} - 32 w - 768 \\ 0 & = (w - 48) (w + 16) \\ w & = 48 or w = - 16 \end{aligned}

$\begin{aligned} 0 & = w^2 - \text{32}w - \text{768}\\ 0 & = (w -\text{48})(w + \text{16})\\ w & = \text{48} \quad \text{or } \quad w = -\text{16} \end{aligned}$

This means that the width of the rectangle is either $\text{48} \ \text{cm}$ or $-\text{16} \ \text{cm}$ . But the dimensions of a rectangle cannot be negative, so $w = \text{48} \ \text{cm}$ .

STEP: Calculate the length of the rectangle
[−1 point ⇒ 0 / 8 points left]

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects $l$ to $w$ .

\begin{aligned} l & = w + 16 \\ = (48) + 16 \\ = 64 cm \end{aligned}

$\begin{aligned} l &= w + \text{16} \\ &= \left(\text{48}\right) + \text{16} \\ & = \text{64}\ \text{cm} \end{aligned}$

The width of the rectangle is $\text{48} \ \text{cm}$ and the length is $\text{64} \ \text{cm}$ .

Submit your answer as: and

Word problems: rectangle facts

The diagonal of a rectangle is $\text{34} \ \text{cm}$ more than its width. The length of the same rectangle is $\text{17}\ \text{cm}$ more than its width.

Determine the width and length of the rectangle.

Answer: The width is

cm

$\text{cm}$ and length is

cm

$\text{cm}$ .

numeric

HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.

STEP: Draw a diagram and label it
[−1 point ⇒ 7 / 8 points left]

Notice we have added labels for the three parts of the rectangle mentioned in the question: $d$ is the diagonal, $w$ is the width, and $l$ is the length.

STEP: Connect the variables in the diagram
[−2 points ⇒ 5 / 8 points left]

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

\begin{array}{lcl} _{34 cm more than its width}^{the diagonal of the rectangle is} & \to d & = w + 34 \\ _{17 cm more than its width}^{the length of the rectangle is} & \to l & = w + 17 \end{array}

$\begin{array}{lcl} ^{\text{the diagonal of the rectangle is}}_{ \text{34} \ \text{cm} \text{ more than its width}} & \rightarrow \quad d &= w + \text{34} \\ \\ ^{\text{the length of the rectangle is}}_{\text{17}\ \text{cm} \text{ more than its width}} & \rightarrow \quad l &= w + \text{17} \end{array}$

STEP: Use the Theorem of Pythagoras and solve the equations simultaneously
[−2 points ⇒ 3 / 8 points left]

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

\begin{aligned} d^{2} & = w^{2} + l^{2} \end{aligned}

$\begin{aligned} d^2 &= w^2 + l^2 \end{aligned}$

And now we can substitute in the equations from above. Specifically we can substitute $w + \text{34}$ in for $d$ and $w + \text{17}$ in for $l$ . Then expand the binomials and simplify the equation as much as possible.

\begin{aligned} d^{2} & = w^{2} + l^{2} \\ (w + 34)^{2} & = w^{2} + (w + 17)^{2} \\ w^{2} + 68 w + 1 156 & = w^{2} + (w^{2} + 34 w + 289) \\ 0 & = w^{2} - 34 w - 867 \end{aligned}

$\begin{aligned} d^2 &= w^2 + l^2 \\ (w + \text{34})^2 & = w^2 + (w + \text{17})^2\\ w^2 + \text{68}w + \text{1 156} & = w^2 + (w^2 + \text{34}w + \text{289})\\ 0 & = w^2 - \text{34}w - \text{867} \end{aligned}$

STEP: Solve the equation for

w

$w$
[−2 points ⇒ 1 / 8 points left]

We have a quadratic equation in standard form! It is time to solve it.

\begin{aligned} 0 & = w^{2} - 34 w - 867 \\ 0 & = (w - 51) (w + 17) \\ w & = 51 or w = - 17 \end{aligned}

$\begin{aligned} 0 & = w^2 - \text{34}w - \text{867}\\ 0 & = (w -\text{51})(w + \text{17})\\ w & = \text{51} \quad \text{or } \quad w = -\text{17} \end{aligned}$

This means that the width of the rectangle is either $\text{51} \ \text{cm}$ or $-\text{17} \ \text{cm}$ . But the dimensions of a rectangle cannot be negative, so $w = \text{51} \ \text{cm}$ .

STEP: Calculate the length of the rectangle
[−1 point ⇒ 0 / 8 points left]

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects $l$ to $w$ .

\begin{aligned} l & = w + 17 \\ = (51) + 17 \\ = 68 cm \end{aligned}

$\begin{aligned} l &= w + \text{17} \\ &= \left(\text{51}\right) + \text{17} \\ & = \text{68}\ \text{cm} \end{aligned}$

The width of the rectangle is $\text{51} \ \text{cm}$ and the length is $\text{68} \ \text{cm}$ .

Submit your answer as: and

Word problems: rectangle facts

The diagonal of a rectangle is $\text{32} \ \text{cm}$ more than its width. The length of the same rectangle is $\text{16}\ \text{cm}$ more than its width.

What are the width and the length of the rectangle?

Answer: The width is

cm

$\text{cm}$ and length is

cm

$\text{cm}$ .

numeric

HINT: <no title>
[−0 points ⇒ 8 / 8 points left]

The question is about a rectangle. Start by drawing a picture of a rectangle, and then label everything you know about it.

STEP: Draw a diagram and label it
[−1 point ⇒ 7 / 8 points left]

Notice we have added labels for the three parts of the rectangle mentioned in the question: $d$ is the diagonal, $w$ is the width, and $l$ is the length.

STEP: Connect the variables in the diagram
[−2 points ⇒ 5 / 8 points left]

Now we need to connect the information given in the question to the diagram. We need to use the facts from the question to write equations which link the three variables together.

\begin{array}{lcl} _{32 cm more than its width}^{the diagonal of the rectangle is} & \to d & = w + 32 \\ _{16 cm more than its width}^{the length of the rectangle is} & \to l & = w + 16 \end{array}

STEP: Use the Theorem of Pythagoras and solve the equations simultaneously
[−2 points ⇒ 3 / 8 points left]

It would be nice to solve the equations above simultaneously. But that is not possible yet because the equations include three different variables. That means we need another equation.

\begin{aligned} d^{2} & = w^{2} + l^{2} \end{aligned}

$\begin{aligned} d^2 &= w^2 + l^2 \end{aligned}$

\begin{aligned} d^{2} & = w^{2} + l^{2} \\ (w + 32)^{2} & = w^{2} + (w + 16)^{2} \\ w^{2} + 64 w + 1 024 & = w^{2} + (w^{2} + 32 w + 256) \\ 0 & = w^{2} - 32 w - 768 \end{aligned}

STEP: Solve the equation for

w

$w$
[−2 points ⇒ 1 / 8 points left]

We have a quadratic equation in standard form! It is time to solve it.

\begin{aligned} 0 & = w^{2} - 32 w - 768 \\ 0 & = (w - 48) (w + 16) \\ w & = 48 or w = - 16 \end{aligned}

$\begin{aligned} 0 & = w^2 - \text{32}w - \text{768}\\ 0 & = (w -\text{48})(w + \text{16})\\ w & = \text{48} \quad \text{or } \quad w = -\text{16} \end{aligned}$

This means that the width of the rectangle is either $\text{48} \ \text{cm}$ or $-\text{16} \ \text{cm}$ . But the dimensions of a rectangle cannot be negative, so $w = \text{48} \ \text{cm}$ .

STEP: Calculate the length of the rectangle
[−1 point ⇒ 0 / 8 points left]

Finally, we can use the width to calculate the length of the rectangle. Use the equation from above which connects $l$ to $w$ .

\begin{aligned} l & = w + 16 \\ = (48) + 16 \\ = 64 cm \end{aligned}

$\begin{aligned} l &= w + \text{16} \\ &= \left(\text{48}\right) + \text{16} \\ & = \text{64}\ \text{cm} \end{aligned}$

The width of the rectangle is $\text{48} \ \text{cm}$ and the length is $\text{64} \ \text{cm}$ .

Submit your answer as: and

Setting up simultaneous equations

Last week, Abodunrin and Babalwe had a chemistry test. Now they are comparing their marks and they notice these facts:

The sum of the marks is $\text{152}$ .
Abodunrin's mark is $\text{22}$ more than Babalwe's mark.

Let $a$ represent Abodunrin's mark and $b$ represent Babalwe's mark. Then which equations below accurately represent the facts? Select your answer from the choices below.

Answer:

Fact about the test scores	Equation
The sum of the marks is $\text{152}$ .
Abodunrin's mark is $\text{22}$ more than Babalwe's mark.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is more. Use these key words to change the words into operations, for example, addition or multiplication.

STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

Word	Meaning
sum	+
product	$\times$
is	=
consecutive	1 apart
more than	add to
less than	subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable $a$ for Abodunrin's mark and $b$ for Babalwe's mark. So we can break up the first fact like this:

\begin{array}{ccc} \underset{⏟}{The sum of the marks} & is & \underset{⏟}{152} \\ a + b & = & 152 \end{array}

$\begin{array}{ccc} \underbrace{\text{The sum of the marks}} & \text{is} & \underbrace{\quad \text{152}\quad} \\ a + b & = & 152 \end{array}$

The first fact is equivalent to this equation: $a + b = 152$ .

STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

Similarly, we can identify key parts of the second fact, and translate each into an expression.

\begin{matrix} \underset{⏟}{Abodunrin's mark} & is & \underset{⏟}{22 more than Babalwe's mark} \\ a & = & b + 22 \end{matrix}

$\begin{array}{c} \underbrace{\text{Abodunrin's mark}} & \text{is} & \underbrace{\text{22 more than Babalwe's mark}} \\ a & = & b + 22 \end{array}$

This equation, $a = b + 22$ , means that Abodunrin's mark is more that Babalwe's mark.

The correct answers are:

Fact about the test scores	Equation
The sum of the marks is $\text{152}$ .	$a + b = 152$
Abodunrin's mark is $\text{22}$ more than Babalwe's mark.	$a = b + 22$

Submit your answer as: and

Setting up simultaneous equations

Last week, Abdoul and Balarabe had a maths test. Now they are comparing their marks and they notice these facts:

The sum of the marks is $\text{126}$ .
Abdoul's mark is $\text{6}$ more than Balarabe's mark.

Let $a$ represent Abdoul's mark and $b$ represent Balarabe's mark. Then which equations below accurately represent the facts? Select your answer from the choices below.

Answer:

Fact about the test scores	Equation
The sum of the marks is $\text{126}$ .
Abdoul's mark is $\text{6}$ more than Balarabe's mark.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is more. Use these key words to change the words into operations, for example, addition or multiplication.

STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

Word	Meaning
sum	+
product	$\times$
is	=
consecutive	1 apart
more than	add to
less than	subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable $a$ for Abdoul's mark and $b$ for Balarabe's mark. So we can break up the first fact like this:

\begin{array}{ccc} \underset{⏟}{The sum of the marks} & is & \underset{⏟}{126} \\ a + b & = & 126 \end{array}

$\begin{array}{ccc} \underbrace{\text{The sum of the marks}} & \text{is} & \underbrace{\quad \text{126}\quad} \\ a + b & = & 126 \end{array}$

The first fact is equivalent to this equation: $a + b = 126$ .

STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

Similarly, we can identify key parts of the second fact, and translate each into an expression.

\begin{matrix} \underset{⏟}{Abdoul's mark} & is & \underset{⏟}{6 more than Balarabe's mark} \\ a & = & b + 6 \end{matrix}

$\begin{array}{c} \underbrace{\text{Abdoul's mark}} & \text{is} & \underbrace{\text{6 more than Balarabe's mark}} \\ a & = & b + 6 \end{array}$

This equation, $a = b + 6$ , means that Abdoul's mark is more that Balarabe's mark.

The correct answers are:

Fact about the test scores	Equation
The sum of the marks is $\text{126}$ .	$a + b = 126$
Abdoul's mark is $\text{6}$ more than Balarabe's mark.	$a = b + 6$

Submit your answer as: and

Setting up simultaneous equations

Last week, Adanna and Bokamoso had a physics test. Now they are comparing their marks and they notice these facts:

The sum of the marks is $\text{159}$ .
Adanna's mark is $\text{7}$ more than Bokamoso's mark.

Let $a$ represent Adanna's mark and $b$ represent Bokamoso's mark. Then which equations below accurately represent the facts? Select your answer from the choices below.

Answer:

Fact about the test scores	Equation
The sum of the marks is $\text{159}$ .
Adanna's mark is $\text{7}$ more than Bokamoso's mark.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

For the first equation, the key word is sum. For the second equation the key word is more. Use these key words to change the words into operations, for example, addition or multiplication.

STEP: Translate the first fact into an equation
[−1 point ⇒ 1 / 2 points left]

Word	Meaning
sum	+
product	$\times$
is	=
consecutive	1 apart
more than	add to
less than	subtract from

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question tells us that we should use the variable $a$ for Adanna's mark and $b$ for Bokamoso's mark. So we can break up the first fact like this:

\begin{array}{ccc} \underset{⏟}{The sum of the marks} & is & \underset{⏟}{159} \\ a + b & = & 159 \end{array}

$\begin{array}{ccc} \underbrace{\text{The sum of the marks}} & \text{is} & \underbrace{\quad \text{159}\quad} \\ a + b & = & 159 \end{array}$

The first fact is equivalent to this equation: $a + b = 159$ .

STEP: Translate the second fact into an equation
[−1 point ⇒ 0 / 2 points left]

Similarly, we can identify key parts of the second fact, and translate each into an expression.

\begin{matrix} \underset{⏟}{Adanna's mark} & is & \underset{⏟}{7 more than Bokamoso's mark} \\ a & = & b + 7 \end{matrix}

$\begin{array}{c} \underbrace{\text{Adanna's mark}} & \text{is} & \underbrace{\text{7 more than Bokamoso's mark}} \\ a & = & b + 7 \end{array}$

This equation, $a = b + 7$ , means that Adanna's mark is more that Bokamoso's mark.

The correct answers are:

Fact about the test scores	Equation
The sum of the marks is $\text{159}$ .	$a + b = 159$
Adanna's mark is $\text{7}$ more than Bokamoso's mark.	$a = b + 7$

Submit your answer as: and

Apply linear & quadratic equations

1. Revision of solving equations

Exercises

2. Revision of simultaneous equations

Points of intersection

Simultaneous equations: solving by substitution

Simultaneous equations

Simultaneous equations

Simultaneous equations

Simultaneous equations: intersection points

Substitution with quadratic equations

Equations with fractions

Simultaneous equations: special outcomes

Simultaneous equations

Simultaneous equations with xyxyxy terms

Exercises

Points of intersection

Points of intersection

Points of intersection

Simultaneous equations: solving by substitution

Simultaneous equations: solving by substitution

Simultaneous equations: solving by substitution

Simultaneous equations

Simultaneous equations

Simultaneous equations

Simultaneous equations

Simultaneous equations

Simultaneous equations

Simultaneous equations

Simultaneous equations

Simultaneous equations

Simultaneous equations: intersection points

Simultaneous equations: intersection points

Simultaneous equations: intersection points

Substitution with quadratic equations

Substitution with quadratic equations

Substitution with quadratic equations

Equations with fractions

Equations with fractions

Equations with fractions

Simultaneous equations: special outcomes

Simultaneous equations: special outcomes

Simultaneous equations: special outcomes

Simultaneous equations

Simultaneous equations

Simultaneous equations

Simultaneous equations with xyxyxy terms

Simultaneous equations with xyxyxy terms

Simultaneous equations with xyxyxy terms

3. Practical applications

Setting up simultaneous equations

Word problems: distance, speed and time

Word problems: solving simultaneous equations

Word problems: checking answers about shopping

Word problems: checking answers with shapes

Word problems: test scores and simultaneous equations

Word problems: odd and even numbers

Word problems: checking answers

Word problems: products of odd and even numbers

Word problems: finding consecutive numbers

Word problems: an age-old question

Word Problems: checking answers

Word problems: rectangle facts

Setting up simultaneous equations

Exercises

Setting up simultaneous equations

Setting up simultaneous equations

Setting up simultaneous equations

Word problems: distance, speed and time

Word problems: distance, speed and time

Word problems: distance, speed and time

Word problems: solving simultaneous equations

Word problems: solving simultaneous equations

Word problems: solving simultaneous equations

Word problems: checking answers about shopping

Word problems: checking answers about shopping

Word problems: checking answers about shopping

Word problems: checking answers with shapes

Word problems: checking answers with shapes

Word problems: checking answers with shapes

Simultaneous equations with $xy$ terms

Simultaneous equations with $xy$ terms

Simultaneous equations with $xy$ terms

Simultaneous equations with $xy$ terms